Answer Key
Conic Sections Answers
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University:
Wilkes University -
Course:
MTH 343 | Geometry Academic year:
2023
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421
Pages:
14
Author:
gewoekerdodx
Page # 1
CONIC SECTIONS
CONIC SECTIONS
CIRCLE
A circle is the locus of a point which moves in a
plane in such a way that its distance from a fixed
point is always constant. The fixed point is called
the centre and constant distance is called the
radius of the circle.
1. STANDARD FORMS OF EQUATION OF A
CIRCLE
2.1 General Equation of a Circle –
The general equation of a circle is
x2 + y2 + 2gx + 2fy + c = 0, Where g, f, c
NOTE :
(i)
If the centre is origin, then the equation o f
the circle is x2 + y2 = r2
(ii) If r = 0 than circle is called point circle and its
equation is
(x – h)2 + (y – k)2 = 0
Diametral Form – If (x1, y1) and
(x2, y2) be the extremities of a diameter, then the
equation of circle is (x – x1)
(x – x2) + (y – y1) (y – y2) = 0
2. EQUATION OF A CIRCLE IN SOME SPECIAL
are constants.
CASES
(i)
(i)
Centre of a general equation of a circle is
(–g, –f ) i.e. (–
1
coefficient of
2
1
coefficient of y)
2
(ii) Radius of a general equation of a circle is
x, –
g2 f 2 c
NOTE :
The general equation of second degree
2
ax + by2 + 2hxy + 2gx + 2fy + c = 0 represents
If centre of circle is (h, k) and passes through
origin then its equation is
(x – h)2 + (y – k)2 = h2 + k2
x2 + y2 – 2hx – 2ky = 0
(iii) If the circle touches x axis then its equation
is (Four cases)
(x ± h)2 + (y ± k)2 = k2
(i)
a circle if a = b 0 and h = 0.
(ii) Locus of a point P represent a circle if its
distance from two points A and B is not equal i.e.
PA = kPB represent a circle if k 1
(ii) General equation of a circle represents –
(a) A real circle if g2 + f 2 – c > 0
(b) A point circle if g2 + f 2 – c = 0
(c) An imaginary circle if g2 + f 2 – c < 0
(iv)In General equation of a circle –
(a) If c = 0 The circle passes through origin
(b) If f = 0 The centre is on x – axis
(c) If g = 0 The centre is on y– axis
Y
(–h,k)
(h,k)
K
K
O
K
(–h,–k)
(iii) If the circle touches y axis then its equation
is (Four cases)
Y
h
(–h,k)
h
(h,k)
(h, k) and radius r is
(x – h)2 + (y – k)2 = r2
r
(h, k)
(x, y)
1
X
O
h
(–h,–k)
h
(h,–k)
(x ± h)2+(y ± k)2 = h2
(iv) If the circle touches both the axis then its
equation is (Four cases) (x ± r)2 + (y ± r)2 = r2
Central Form of Equation of a circle The equation of a circle having centre
X
K
(h,–k)
Y
(r,r)
(–r,r)
X
O
(–r,–r)
(r,–r)
(v) If the circle touches x axis at origin (Two
cases) x2 + (y ± k)2 = k2
x2 + y2 ± 2ky = 0 Page # 2
CONIC SECTIONS
perpendicular from the centre of circle on the line,
then if
Y
B
(0,k)
p
X
(0,-k)
r
A
(vi) If the circle touches y axis at origin (Two
cases)
(x ± h)2 + y2 = h2
x2 + y2 ± 2xh = 0
Y
(-h,0)
(h,0)
X
p>r
p=r
p 0 Point is outside the circle.
S1 = 0 Point is on the circle.
S1 < 0 Point is inside the circle.
4. The least and greatest distance of a point from
a circle
Let S = 0 be a circle and A (x1, y1) be a point. If
the diameter of the circle which is passing through
the circle at P and Q. then
Q
r
P
1 m2
C
A(x1, y1)
AP = AC – r = least distance
AQ = AC + r = greatest distance,
where ‘r’ is the radius and C is the centre of
circle
5. LINE AND CIRCLE
Let L = 0 be a line and S = 0 be a circle, if ‘r’ be
the radius of a circle and p be the length of
1. DEFINITION
A parabola is the locus of a point which moves
in such a way that its distance from a fixed
point is equal to its perpendicular distance from
a fixed straight line.
1.1 Focus : The fixed point is called the focus
of the Parabola.
1.2 Directrix : The fixed line is called the
directrix of the Parabola.
2. TERMS RELATED TO PARABOLA
2.1 Eccentricity :
If P be a point on the
parabola and PM and PS are the distances from
the directrix and focus S respectively then the
ratio PS/PM is called the eccentricity of the
Parabola which is denoted by e.
Note:By the definition for the parabola e = 1.
If e > 1 Hyperbola,
e = 0
circle,
e < 1
ellipse
2.2 Axis : A straight line passes through the
focus and perpendicular to the directrix is called
the axis of parabola.
2.3 Vertex : The point of intersection of a
parabola and its axis is called the vertex of the
Parabola. Page # 3
CONIC SECTIONS
NOTE:
The vertex is the middle point of the
focus and the point of intersection of axis and
directrix
2.4 Focal Length (Focal distance) : The
distance of any point P (x, y) on the parabola
from the focus is called the focal length. i.e.
The focal distance of P = the perpendicular
distance of the point P from the directrix.
2.5 Double ordinate : The chord which is
perpendicular to the axis of Parabola or parallel
to Directrix is called double ordinate of the
Parabola.
2.6 Focal chord : Any chord of the parabola
passing through the focus is called Focal chord.
2.7 Latus Rectum : If a double ordinate passes
through the focus of parabola then it is called
as latus rectum.
2.7.1Length of latus rectum :
The l ength of the l atus rectum = 2 x
perpendicular distance of focus from the
directrix.
3. STANDARD FORM OF EQUATION OF PARABOLA
If we take vertex as the origin, axis as x- axis
and distance between vertex and focus as 'a'
then equation of the parabola in the simplest
form will bey2 = 4ax
3.1Parameters of the Parabola y2 = 4ax
(i) Vertex A (0, 0)
(ii) Focus S (a, 0)
(iii) Directrix x + a = 0
(iv) Axis y = 0 or x– axis
(v) Equation of Latus Rectum x = a
(vi) Length of L.R. 4a
(vii) Ends of L.R. (a, 2a), (a, – 2a)
(viii) The focal distance sum of abscissa of
the point and distance between vertex and L.R.
(ix) If length of any double ordinate of parabola
y2 = 4ax is 2 then coordinates of end points
of this Double ordinate are
2
2
and , .
,
4a
4a
4. REDUCTION TO STANDARD EQUATION
If the equation of a parabola is not in standard
form and if it contains second degree term
either in y or in x (but not in both) then it can
be reduced into standard form. For this we
change the given equation into the following
forms(y–k)2 = 4a(x–h) or (x–p)2 = 4b(y–q)
1
And then we compare from the following table
for the results related to parabola.
5. GENERAL EQUATION OF A PARABOLA
If (h,k) be the locus of a parabola and the
equation of directrix is ax + by + c = 0,
then its equation is given by
(x h ) 2 ( y k ) 2
=
ax by c
a2 b2
which gives
(bx– ay)2 + 2gx + 2fy + d = 0 where g, f, d are
the constants.
Note:
(i) It is a second degree equation in x and
y and the terms of second degree forms a
perfect square and it contains at least one
linear term.
(ii) The general equation of second degree
ax 2 + by 2 + 2hxy + 2gx + 2fy + c = 0
represents a
parabola, if
(a) h2 = ab
(b) = abc + 2fgh – af2 – bg2 –ch2 0
6. EQUATION OF PARABOLA WHEN ITS VERTEX AND
FOCUS ARE GIVEN
6.1 If both lie on either of the coordinate
axis :
In this case first find distance 'a' between
these points and taking vertex as the origin
suppose the equation as y2 = 4ax or
x2 = 4ay. Then shift the origin to the vertex.
6.2 When both do not lie on any coordinate
axes
In this case first find the coordinates of Z and
equation of the directrix, then write the equation
of the parabola by the definition.
ELLIPSE
1
DEFINITION
An ellipse is the locus of a point which moves
in such a way that its distance form a fixed
point is in constant ratio to its distance from
a fixed line. The fixed point is called the focus
and fixed line is called the directrix and the
constant ratio is called the eccentricity of a
ellipse denoted by (e).
In other word, we can say an ellipse is the
locus of a point which moves in a plane so that
the sum of it distances from fixed points is Page # 4
CONIC SECTIONS
constant.
2. EQUATION OF AN ELLIPSE
2.1 Standard Form of the equation of ellipse
Let the distance between two fixed points S
and S' be 2ae and let C be the mid point of SS.
Taking CS as x- axis, C as origin.
Let P(h,k) be the moving point Let SP+ SP =
2a (fixed distance) then
SP+S'P
= {(h - ae)2 + k 2 } + {(h + ae)2 + k 2 } = 2a
h2(1– e2) + k2 = a2(1– e2)
Hence Locus of P(h, k) is given by.
x2(1– e2) + y2 = a2(1– e2)
x2
a2
+
y2
2
a (1 e 2 )
= 1
y
P(h,k)
C
x
S(ae,0)
S'(–ae,0)
together are called the principal axes and its
length will be given by
Length of major axis 2a
Length of minor axis 2b
(iii) Centre : The point which bisect each chord
of an ellipse is called centre (0,0) denoted by 'C'.
(iv) Directrix : ZM and Z M are two directrix
and their equation are x= a/e and x = – a/e.
(v) Focus : S (ae, 0) and S (–ae,0) are two
foci of an ellipse.
(vi) Latus Rectum : Such chord which passes
through either focus and perpendicular to the
major axis is called its latus rectum.
Length of Latus Rectum :
If L i s (ae,) then 2 i s t he l ength of
Latus Rectum.
2b 2
Length of Latus rectum is given by
.
a
(vii) Relation between constant
a, b, and e
b2 = a2(1– e2) e2 =
Let us assume that a2(1– e2 )= b2
The standard equation will be given by
x2
a2
y2
b2
1
2.1.1 Various parameter related with
standard ellipse :
Let the equation of the ellipse
a2
y2
b2
X'
L1
a
e
M
L
Z' (–a,0) A'
S' (–ae,0)
C
Directrix
x
Y
B(0,b) P(x,y)
M'
L1'
L'
B'(0,–b)
S (ae,0)
a2
a 2 b2
e =
a2
Result :
(a) Centre C is the point of intersection of the
axes of an ellipse. Also C is the mid point of AA.
(b) Another form of standard equation of ellipse
x2
a2
y2
b2
1 when a < b.
In this case major axis is BB= 2b which is
along y- axis and minor axis is AA= 2a along
x- axis. Focus S(0,be) and S(0,–be) and
directrix are y = b/e and y = –b/e.
1 a > b
x
Z
A(a,0)
a
e
X
Directrix
x2
a 2 b2
Y'
(i) Vertices of an ellipse : The point of which
ellipse cut the axis x-axis at A(a,0) & A’(– a,
0) and y- axis at B(0, b) &
B’(0, – b) is called the vertices of an ellipse.
(ii) Major & Minor axis : The straight line AA
is called major axis and BB is called minor axis.
The
major
a nd
minor
axi s
taken
2.2 General equation of the ellipse
The general equation of an ellipse whose focus
is (h,k) and the directrix is the line ax + by + c
= 0 and the eccentricity will be e. Then let
P(x1,y1) be any point on the ellipse which moves
such that SP = ePM
(x1–h)2 + (y1–k)2
2
=
2
e (ax1 + by1 + c)
2
2
a +b
Hence the locus of (x1,y1) will be given by
(a2 + b2) [(x – h)2 + (y–k)2]
= e2(ax + by + c)2
Which is the equation of second degree from Page # 5
CONIC SECTIONS
which we can say that any equation of second
degree represent equation of an ellipse.
Note : Condition for second degree in X & Y to
represent an ellipse is that if
axes
axes
2
h2 = ab < 0 & = abc + 2
(vi) Focal distance : The distance of any point
fgh – af – bg – ch 0
on the hyperbola from the focus is called the
2
2
2
focal distance of the point.
HYPERBOLA
Note :
1. STANDARD EQUATION AND DEFINITIONS
Standard Equation of hyperbola is
z’
y
x2
a2
–
y2
b2
= 1
z
M’
S’
K’
A’
(–ae, 0)
(–A, 0)
|SP – SP| = 2a (const.)
The hyperbola whose transverse and conjugate
P(x,y)
L
K
A
S
(ae,0)
(a,0)
O
equal to the length of the transverse axes.
2. CONJUGATE HYPERBOLA
M
(0,0)
The difference of the focal distance of
a point on the hyperbola is constant and is
T
–x
2
2b
Conjugate
= 1+
e = 1
Transverse
2a
x
L’
T’
axes are respectively the conjugate and
transverse axes of a given hyperbola is called
conjugate hyperbola.
(i) Definition hyperbola : A Hyperbola is the
Note :
locus of a point in a plane which moves in the
(i) If e1 and e2 are the eccentricities of the
plane in such a way that the ratio of its distance
hyperbola and its conjugate then
from a fixed point (called focus) in the same
1
e1
2
+
1
e22
= 1
plane to its distance from a fixed line (called
(ii) The focus of hyperbola and its conjugate
directrix) is always constant which is always
are concyclic.
greater than unity.
(ii) Vertices : The point A and A where the
Length of transverse axis :
curve meets the line joining the foci S and S
The length of transverse axis = 2b = 8
are called vertices of hyperbola.
(iii) Transverse and Conjugate axes : The
straight line joining the vertices A and A is
Length of conjugate axis :
The length of conjugate axis = 2a = 6
called transverse axes of the hyperbola. Straight
line perpendicular to the transverse axes and
Eccentricity : e = 1
passes through its centre called conjugate axes.
(iv) Latus Rectum : The chord of the hyperbola
whi ch passes th roug h th e focus and is
perpendicular to its transverse
axes is called
2
latus rectum. Length of latus rectum =
(v) Eccentricity : For the hyperbola
x2
a2
–
y2
b2
2b
.
a
a2
b
2
=
1
Foci : the co- ordinates of the foci are (0,±
be), i.e., (0, ± 4)
Length of Latus rectum :
The length of latus rectum
2a 2
9
2 (3) 2
=
=
b
2
4
Equation of directrices :
=
The equation of directrices are
= 1,
b2
=
a2
(e2
– 1)
9
5
=
16
4
y = ±
b
e
y = ±
4
16
5 / 4 = ± 5 Page # 6
CONIC SECTIONS
SOLVED PROBLEMS
CIRCLE
Ex.1
Sol.
Find the area of a circle which passes through
(4, 6) and its centre is (1, 2).
Radius r
= (4 1) 2 (6 2) 2 = 5
Area = r2 = .52 = 25
Ex.2
Sol.
Ex.3
Sol.
Find the centre of the circle
x2 + y2 – 2x + 4y + 1 = 0
2g = – 2
g=–1
2f = 4
f=2
Centre is (1, –2)
Find the radius of the circle
2 (x2 + y2) + 4x – 3y + 1 = 0
First making the coefficient of x2 and
y2, 1 by dividing the equation with 2
3
1
x2 +y2 + 2x – y + = 0
2
2
2g = 2
g=1
3
2f = –
2
c = 1/2
3
f=–
4
2
1
17
17
2 3
r = (1) =
=
2
16
4
4
Ex.4
Sol.
If (6, –3) is the one extremity of diameter to
the circle x2 + y2 – 3x + 8y – 4 = 0 then find
its other extremity
Centre of circle is (3/2, –4)
Let the other extremity is (h, k)
Ex.5
Sol.
Ex.6
Sol.
6 h 3 3 k
= ,
= – 4 (–3, –5)
2 2 2
If y = 2x + K is a diameter to the circle
2(x2 + y2) + 3x + 4y –1 = 0, then find K
Centre of circle (–3/4, –1)
this lies on diameter y = 2x + K
–1 = –3/4 × 2 + K
K = 1/2
Find the equation of a circle whose centre is
(2, –1) and radius is 3.
(x – 2)2 + (y + 1)2 = 32
x2 – 4x + 4 + y2 + 2y + 1 = 9
x2 + y2 – 4x + 2y – 4 = 0
1
Ex.7
Sol.
Find the equation of a circle with centre at
the origin and which passes through (7, –2).
Radius
r = 7 2 (2) 2 = 53
Equation of circle is x2 + y2 = 53
Ex.8
Sol.
Ex.9
Sol.
Find the equation of the circle touches y axis
and having centre is (–2, –3)
Here radius of circle | –2 | = 2
Equation is (x + 2)2 + (y + 3)2 = 22
or x2 + y2 + 4x + 6y + 9 = 0
A circle touches x- axis at +3 distance and
cuts an intercept of 8 in +ve direction of yaxis. Then find its equation
From figure.
Y
M 3 c
4
5
A
O T
X
Radius of Circle = 32 4 2 = 5 and
centre
is
(3,5)
Hence equation is (x–3)2 + (y – 5)2 = 52
x2 + y2 – 6x – 10y + 9 = 0
Ex.10 If the line x – y + 1 = 0 is a chord of the circle
x2 + y2 – 2x + 4y – 4 = 0 then find the length
of this chord Sol.
Centre of the circle = (1, –2)
Radius = 1 4 4 = 3
Here p =
1 2 1
2
=2 2
Length of chord =2 a 2 p 2 .
=
9 8 = 2
Ex.11 Find the length of intercept on y axis, by a
circle whose diameter is the line joining the
points (–4, 3) and (12, –1)
Sol.
Here equation of the circle
(x + 4) (x – 12) + (y – 3) (y + 1) = 0
or x2 + y2 – 8x – 2y – 51 = 0
Hence intercept on y axis
=
2 f2 c
=
2 1 (51) = 4 13 Page # 7
CONIC SECTIONS
PARABOLA
Axis is Y = 0
Ex.12 Find the vertex of the parabola
y2 +6x – 2y +13 = 0
Sol.
We have : y2 + 6x – 2y + 13 = 0
y2 – 2y = –6x – 13
(y – 1)2 = – 6(x + 2)
Clearly, the vertex of this parabola is (–2,1)
2y – 1 = 0
Ex.13 If vertex of parabola is (2, 0) and directrix
is y-axis, then find its focus
Sol.
Since the axis of the parabola is the line
wh ich passes through vertex and
perpendicular to the directrix, therefore xaxis is the axis of the parabola.
Obviously Z (0, 0).
Let focus of the parabola is S (a, 0). Since
vertex (2,0) is mid point of
ZS, therefore
a0
= 2 a = 4.
2
Focus is (4, 0)
Ex.14 If the focus of a parabola is (1, 0) and its
directrix is x+y=5, then find its vertex
Sol.
Since axis is a line perpendicular to directrix,
so it will be x – y = k. It also passes from
focus, therefore k = 1.
So equation of axis is x – y = 1.
Solving it with x + y = 5, we get
Z (3, 2).
If vertex is (a, b), then a = 2, b = 1.
Hence vertex is (2, 1).
Ex.15 Find the directrix and axis of the parabola
4y2 – 6x – 4y = 5
Sol.
Here 4y2 – 4y = 6x + 5
1
4y
2
2
= 6 (x + 1)
1
= Y, x + 1 = X
2
The equation in standard form
3
X
2
4a =
3
2
a =
Directrix,
x +1+
3
8
X + a = 0
3
= 0
8
8x + 11 = 0
1
= 0
2
Ex.16 Find the length of latus rectum of a parabola,
whose focus is (2, 3) and directrix is the
line x – 4y + 3 = 0
Sol.
The length of latus rectum
= 2 × perp. from focus to the directrix
= 2 ×
2 4 (3) 3
2
(1) (4)
2
=
14
17
The numerical length =
14
17
Ex.17 Find the coordinates of an endpoint of the
latus rectum of the parabola
(y – 1)2 = 4(x + 1)
Sol.
Shifting the origin at (–1, 1) we have
x X 1
y Y 1
...(i)
Using (i), the given parabola becomes.
Y2 = 4X
The coordinates of the endpoints of latus
rectum are
(X = 1, Y = 2) and (X =1, Y= –2)
Using (i), the coordinates of the end point
of the latus rectum are (0,3) and (0, –1)
Ex.18 If focus of a parabola is (3,–4) and directrix
is x+y–2= 0, then find its vertex
Sol. First we find the equation of axis of parabola,
which is perpendicular to directrix. So its
equation is
x – y + k = 0. It passes through focus S
(3, –4)
3 – (–4) + k = 0 k = –7
Let Z is the point of intersection of axis and
directrix.
Solving equation x + y –2 = 0 and x –y–7=0
gives Z (9/2, –5/2)
Vertex A is the mid point of Z and S
Put y –
Y2 =
y –
3 9 4 5
2
2
A
2 ,
2
=
15 13
A ,
4
4
Ex.19 Find the equation of the parabola whose
vertex is (–3, 0) and directrix is x + 5 = 0
Sol.
A line passing through the vertex
(–3, 0) and perpendicular to directrix x + 5
= 0 is x-axis which is the axis of the parabola
by definition.
Let focus of the parabola is (a, 0). Since
vertex, is the middle point of Z(–5, 0) and Page # 8
CONIC SECTIONS
focus S, therefore
2
(a 5)
a –1
2
Focus = (–1,0)
Thus the equation to the parabola is
(x + 1)2 + y2 =(x + 5)2
y2 = 8(x +3)
3
Ex.20 Find the equation of the directrix of the
parabola y2 = 12x
Sol.
Here a = 3, so the equation of the directrix
is given by
x = – a x= – 3
x + 3 = 0
Ex.21 Find the equation of the latus rectum of the
parabola x2 = – 12y
Sol.
Here a = 3 so the equation of the L.R. is
given by y = – a y = – 3
Ex.22 Find the axis of the parabola
9y2 –16x –12y –57 = 0
Sol.
9y2 –12y = 16x + 57
(3y – 2)2 = 16x + 57
2
2
16
61
y x
3
9
16
Which shows that equation of the axis is
y –2/3 = 0 or 3y = 2
Ex.23 Find the Vertex, focus, latus rectum, length
of the latus rectum and equation of directrix
of the parabola y2 = 4x +4y
Sol. Given parabola y2 = 4x + 4y
or (y – 2)2 =4(x +1)
or Y2 = 4X
Here X = x +1, Y= y – 2
vertex = (X = 0, Y = 0)
or (x + 1 = 0, y – 2 = 0) (–1, 2)
Focus (X = 1, Y = 0)
or (x + 1 = 1, y – 2 = 0) (0, 2)
Latus rectum X = 1 x = 0
Length of Latus rectum = 4
equation of the directrix is
X = –1
x +1 = –1
x = –2
Ex.24 Find the vertex of the parabola
x2 – 8y – x + 19 = 0
Sol. The given equation of Parabola can be written
as
1
x
2
2
– 8y + 19 –
2
1
76 1
x = 8y –
2
4
1
1
= 0
4
75
1
x = 8 y 32
2
1 75
vertex = 2 , 32
Ex.25 If (0,4) and (0,2) are vertex and focus of
parabola respectively, then find its equation
Sol.
Since vertex and focus are on y-axis, so yaxis is the axis of the parabola.
Distance between vertex and focus
= 'a' = 2.
So on taking vertex as origin, the equation
of the parabola is x2 = – 4ay
(negative because vertex lies above the
focus)
or x2 = – 8y.
Now shifting the origin to its original position,
the required equation will be
x2 = –8(y – 4)
x2 + 8y = 32
ELLIPSE
Ex.26 Find the equation of the ellipse which passes
through origin and has its foci at the points
(1, 0) and (3, 0)
Sol:
Centre being mid point of the foci is
1 3
,0 = (2, 0)
2
Distance between foci 2ae = 2
ae = 1 or a2 – b2 = 1
If the ellipse
( x 2)
2
2
+
y
...(i)
2
a
b2
passes from (0, 0)
4
= 1 a2 = 4
a2
from (i)
b2 = 3
= 1, then as it
y2
( x 2) 2
+
= 1
3
4
2
2
3x + 4y – 12x = 0
Hence
or
Ex.27 A man running round a racecourse notes
that the sum of the distance of two flag
posts from him is always 10 meters and the
distance between the flag posts is 8 meters.
Find the area of the path he encloses
Sol.: The race course will be an ellipse with the
flag posts as its foci. If a and b are the
semi major and minor axes of the ellipse,
then sum of focal distances 2a = 10 and
2ae = 8
a = 5, e = 4/5 Page # 9
CONIC SECTIONS
16
= 9
= 25 1
25
Area of the ellipse = ab =.5.3 =15
or
Ex.28 Find the equation of the ellipse whose
eccentricity is 1/2, the focus is (–1, 1) and
the directrix is x – y + 3 = 0.
Sol.
Let P (x,y) be any point on the ellipse
whose focus is S(–1,1) and eccentricity e
=1/2. Let PM be perpendicular from P on the
directrix. Then,
X2
+
Y2
= 1
a2
b2
Centre X = 0, Y = 0 i.e. (1,2).
b2 = a2(1 – e2)
Length of major axis = 2a = 2
Length of minor axis = 2b = 2/ 3 and
(a 2 b 2 ) / a = 1/ 3
e =
Ex.31 Find the radius of the circle passing through
1
SP = ePM SP = (PM)
2
x2
+
16
the foci of the ellipse
y2
= 1, and
9
having its centre (0,3)
4 (SP) = PM
2
2
2
x y3
4 [(x +1)2 + (y –1)2] = 2
1 12
8 (x2 + y2 + 2x – 2y + 2) = (x– y+3)2
7x2 + 7y2+ 10x – 10y + 2xy +7 = 0
This is the required equation of the ellipse.
Ex.29 Find the equation of the ellipse whose axes
are along the coordinate axes, vertices are
(± 5,0) and foci at (± 4,0).
Sol.
Let the equation of the required ellipse be
x
2
a
2
+
y
b2
= 1
x2
y2
, we get
+
= 1,
25
9
which is the equation of the required ellipse.
Ex.30 Find the centre, the length of the axes and
the eccentricity of the ellipse
2x2+3y2–4x–12y+13 = 0.
Sol.
The given equation can be rewritten as
2[x2 – 2x] + 3 [y2 – 4y] + 13 = 0
or 2 (x – 1)2 + 3 (y– 2)2 = 1
(1 / 2 ) 2
+
( y 2) 2
(1 / 3 ) 2
= 1,
b2
a2
= 1 –
9
e =
16
7
4
Foci are (± ae, 0) or (± 7 ,0).
Centre is (0,3)
Radius = 7 9 = 4
or
X2 Y 2
X2
Y2
+
= 1. Form 2 + 2 = 1
16
25
b
a
Major axis lies along y- axis. ;
e2 = 1 –
..(1)
16
= 9.
b2 = 25 1
25
Substituting the values of a2 and b2 in (1)
( x 1) 2
e = 1 –
Ex.32 Fi nd t he eccen tri ci ty of the el li pse
represented by the equation
25x2 + 16y2 – 150 x – 175 = 0
Sol.
25(x2 – 6x + 9) + 16y2 = 175 + 225
or 25(x – 3)2 + 16y2 = 400
2
The coordinates of its vertices and foci are
(± a, 0) and (± ae,0) respectively.
a = 5 and ae = 4 e = 4/5.
Now, b2 = a2 (1– e2)
or
Sol.
b2
a
2
= 1 –
16
3
; e =
25
5
Ex.33 Find the equation of the ellipse having its
Sol.
centre at the point (2,–3), one focus at
(3,– 3) and one vertex at (4,–3).
C (2, – 3), S (3, – 3) and
A (4, – 3)
Now CA = ( 4 2) 2 ( 3 3) 2 = 2
a = 2
Again CS =
( 3 2 ) 2 ( 3 3) 2 = 1
ae = 1 e =
1
1
=
a
2
Let the directrix cut the major- axis at Q.
Then
AS
1
= e =
AQ
2
If Q (, ), then
SA : AQ = e : 1 = 1 : 2 Page # 10
CONIC SECTIONS
6 6
A
,
= (4, – 3)
3
3
6
6
= 4,
= – 3
3
3
= 6, = – 3
Slope of CA = 0, therefore directrix will be
parallel to y- axis.
Since directrix is parallel to y- axis and it
passes through Q(6, – 3)
equation of the directrix is x = 6.
Let P(x,y) be any point on the ellipse, then
1
PS
e= =
=
2 PM
x - 3
2
+ y + 3
2
x-6
2
1
1
=
4
x 32 y 32
x 6 2
;
Ex.35 Find the equation of an ellipse whose focus
is (– 1,1), eccentricity is 1/2 and the
directrix is x – y + 3 = 0.
Sol.
Let P (x,y) be any point on the ellipse
whose focus is S (–1,1) and the directrix is
x – y + 3 = 0.
PM perpendicular from P (x,y) on the directrix
x –y + 3 = 0.
Then by definition
SP = ePM
(SP)2=e2 (PM)2
8 (x2 + y2 + 2x – 2y + 2)
= x2 + y2 + 9 – 2xy + 6x – 6y
7x2 + 7y2 + 2xy + 10x – 10y + 7 = 0
which is the required equation of the ellipse.
Ex.36 The foci of an ellipse are (± 2, 0) and its
eccentricity is 1/2, find its equation.
Sol.
Let the equation of the ellipse be
x2
a2
y - 2
1
Thus foci are -1, 2 ±
2
y2
b2
= 1, Then coordinates of foci are (±
ae = 2
a×
1
1
= 2 e
2
2
a = 4
We have b2 = a2 (1– e2)
1
b2 = 16 1 = 12
4
Thus, the equation of the ellipse is
2
+
a= 1 .(1)
1/4
3/4
Co- ordinates of centre of the ellipse are
given by x + 1 = 0 and y – 2 = 0
Hence centre of the ellipse is (– 1, 2)
If a and b be the lengths of the semi major
and semi-minor axes, then
a2 = 3/4, b2 = 1/4
Length of major axis = 2a =3,
Length of minor axis = 2b = 1
1
3
a =
, b =
2
2
Since b2 = a2 (1 – e2) 1/4 = 3/4 (1– e2)
1
2
3
e = 2/3
ae =
×
=
2
3
2
Co-ordinates of foci are given by
x + 1 = 0, y – 2 = ± ae
+
ae, 0).
Ex.34 Find the centre, the length of the axes,
eccentricity and the foci of the ellipse.
12x2 + 4y2 + 24 x – 16 y + 25 = 0
Sol.
The given equation can be written in the
form 12(x + 1)2 + 4(y – 2)2 = 3
2
2
or (x – 6)2 = 4 [(x– 3)2 + (y + 3)2]
or x2–12x+36 =
4 [x2 – 6x + 9 +y2 + 6y + 9]
or 3x2+ 4y2 – 12x + 24y + 36 = 0
x +1
1 x y 3
(x + 1)2 +(y – 1)2 =
4
2
x 2 y2
= 1
16 12
HYPERBOLA
Ex.37 If the foci of the ellipse
y2
x2
+ 2 = 1 and
16
b
1
y2
x2
–
=
coincide, then find
25
144 81
the value of b2
hyperbola
Sol.
1
y2
x2
–
=
25
144 81
2
b
81
225
e2 = 1+ 2 =1+
=
144
144
a
15
5
e =
=
i.e., e > 1
12
4
Hence the foci are
For hyperbola Page # 11
CONIC SECTIONS
12 5
(±ae, 0)
. , 0 = (±3, 0)
5 4
Now the foci coincide therefore for ellipse
ae = 3 or a2e2 = 9
or
b2
a2 1 2 = 9
a
2
a – b2 = 9
or
16 – b2 = 9
b = 7
th e
x2
y2
–
16
9
of
the
conic
represented by x2–y2–4x+4y+16 = 0
Sol.
We have x – y – 4x + 4y + 16 = 0
2
or
(x2 – 4x) – (y2 – 4y) = – 16
or
(x2 – 4x + 4) – (y2 – 4y + 4)
= – 16
or
or
(x – 2)2 – (y – 2)2 = – 16
( x 2) 2
4
2
–
( y 2) 2
42
= – 1
42
–
y2
42
= –1,
Thi s i s rectangul ar h yperbol a, whose
eccentricity is always
b2
a2
= 1 +
42
42
;
Ex.39 The equation
9x2 – 16y2 – 18x + 32y – 151 = 0 represent
a hyperbola then find
(A) The length of the transverse axes
(B) Length of latus rectum
(C) Equation of directrix
We have
9x2 – 16y2 – 18x + 32y – 151 = 0
9(x2 – 2x) – 16(y2 – 2y) = 151
9(x2–2x+1) – 16(y2 – 2y + 1) = 144
1
= 1
x2
a2
–
y2
= 1
b2
where a2 = 16 and b2 = 9 so
The length of the transverse axes
= 2a = 8
The length of the letus rectum
a
2b 2
=
a
2
The equaiton of the directrix
a
e
x – 1 = ±
x =
2.
e = 2
Sol.
= 1
This is of the form
x = ±
where x = X + 2, y = Y + 2
i.e. e2 = 1 +
2
(y - 1)
9
where x = X + 1 and y = Y + 1
=
Shifting the origin at (2, 2), we obtain
x2
–
the axes
eccentrici ty
2
2
(x - 1)
16
Shifting the origin at (1, 1) without rotating
2
Ex.38 Fi nd
9(x – 1)2 – 16(y – 1)2 = 144
16
5
x = ±
21
11
; x = –
5
5
16
5
+ 1 Page # 12
CONIC SECTIONS
EXERCISE
Q.16
CIRCLE
Q.1
Find the equation of a circle with centre (2, 4)
and radius 5
Q.2
Find the centre and radius of (x3)2+(y1)2=9
Q.3
Find the equation of the circle whose centre is
(2,5) and which passes through the point (3, 2).
Q.4
Find the equation of the circle of radius 5 cm,
whose centre lies on the y-axis and which
passes through the point (3, 2).
Q.5
Q.6
Q.7
Q.8
Q.9
Find the equation of the circle whose centre is
(2, 3) and which passes through the intersection of the lines
3x + 2y = 11 and 2x + 3y = 4.
the circle x2 + y2 4x 6y 3 = 0 and with
touches the y-axis.
Q.17
Find the equation of the circle concentric with
the circle x2 + y2 6x + 12y + 15 = 0 and of
double its area.
Q.18
Prove that the centres of the three circles x2 +
y2 4x 6y 12 = 0, x2 + y2 + 2x + 4y 5 = 0
and x2 + y2 10x 16y + 7 = 0 are collinear.
Q.19
Find the equation of a circle passing through
the origin and intercepting lengths a and b on
the axes.
Find the equation of the circle passing through
the point (1, 3)and having its centre at the
point of intersection of the lines x – 2y = 4 and
2x + 5y + 1 = 0.
Q.20
If two diameters of a circle lie along the lines
xy=9 and x2y=7, and the area of the circle
is 38.5 sq cm, find the equation of the circle.
Q.21
Find the equation of the circle circumscribing
the triangle formed by the lines
x + y = 6, 2x + y = 4 and x + 2y = 5.
One end of a diameter of the circle
x2 + y2 6x + 5y 7 = 0 is A(1, 3). Find the
coordinates of the other end of this diameter.
Find the equation of the circle, the coordinates of the end points of one of whose diameter are A(3, 2) and B(2, 5)
The sides of a rectangle are given by the equations x = 2, x = 4, y = 2 and y = 5. Find the
equation of the circle drawn on the diagonal of
this rectangle as its diameter.
Find the equation of the circle concentric with
PARABOLA
Q.22
Find the coordinates of the focus and the
vertex, the equations of the directrix and the
axis, and length of the latus rectum of the
Show that the equation x +y 4x+6y5= 0
represents a circle. Find its centre and radius.
parabola :
(i) y2 = 12 x
(ii) y2 = –8x
Q.11
Show that the equation x2+y23x+3y+10=0
does not represent a circle.
(iii) x2 = 16y
(iv) x2 = –8y
Q.12
Find the equation of the circle which is cir-
Q.10
2
2
Q.23
at the origin and focus at F(2, 0).
cumscribed about the triangle whose vertices
are A(2, 3), B(5, 2) and C(6, 1). Find the
Q.24
Q.14
Q.15
Find the equation of the circle concentric with
the circle x2 + y2 + 4x + 6y + 11 = 0 and
passing through the point P(5, 4).
Show that the points A(1, 0), B(2, 7), C(8, 1)
and D(9, 6) all lie on the same circle. Find the
equation of this circle, its centre and radius.
Find the equation of the circle which passes
through the points (1, 3) and (2, 1), and has
its centre on the line 2x + y 4 = 0.
1
Find the equation of the parabola with focus
F(4, 0) and directrix x = 4.
centre and radius of this circle.
Q.13
Find the equation of the parabola with vertex
Q.25
Find the equation of the parabola with vertex
at the origin and focus F(0, 5).
Q.26
Find the equation of the parabola with vertex
at the origin, passing through the point P(5, 2)
and symmetric with respect to the y-axis.
Q.27
Find the equation of the parabola which is
symmetric about the y-axis and passes through
the point P(2, 3). Page # 13
CONIC SECTIONS
Q.38
ELLIPSE
Q.28
at (0, 5) and foci at (0, 8).
Find the
(i) lengths of the major and minor axes,
(ii) coordinates of the vertices
(iii) coordinates of the foci
(iv) eccentricity
2
(v) length of the latus rectum of
Q.39
the length 10.
x
y
1
25 9
Q.30
Find the equation of the ellipse the ends of
whose major and minor axes are (4, 0) and
(0, 3) respectively.
Q.32
Q.40
length 8. Also, find its eccentricity.
Q.41
Find the equation of the ellipse whose foci are
Q.34
Q.35
Find the equation of the ellipse with centre at
the origin, major axis on the x-axis and passing
through the points (4, 3) and (1, 4).
Find the equation of the ellipse which passes
through the point (4, 1) and having its foci at
(3, 0).
Q.42
Q.43
Q.44
Q.45
Find the equation of the hyperbola with vertices
at ( 6, 0) and foci at ( 8, 0).
Find the equation of the hyperbola, the length
Find the equation of the hyperbola with
eccentricity
2 and the distance between
whose foci is 16.
Q.46
Find the equation of the hyperbola whose foci
are (0, 13) and the length whose conjugate
axis is 24.
Q.47
Find the equation of the hyperbola whose foci
are (0, 10) and the length of whose latus
rectum is 9 units.
(iv) eccentricity
Q.37
5
.
3
is 3.
(iii) coordinates of the foci
x2 y 2
1
(v) length of the latus Rectum of
9 16
5 , 0) and the eccentricity is
of whose latus rectum is 4 and the eccentricity
Find the
(ii) coordinates of the vertices
Find the equation of the hyperbola whose foci
are (
HYPERBOLA
(i) length of the axes
Find the equation of the hyperbola whose
vertices are ( 2, 0) and the eccentricity is 2.
2
eccentricity is
, the latus rectum is 5 and
3
Q.36
5 , 0) and the length of the latus
rectum is 8 units.
Fi nd the equation of a n ell i pse whose
the centre is at the origin.
Find the equation of the hyperbola whose foci
are (3
Find the equation of the ellipse whose foci are
1
.
2
Find the equation of the hyperbola whose foci
are (5, 0) and the conjugate axis is of the
4
at (0, 4) and e =
.
5
Q.33
29 , 0) and the transverse axis is of
2
Find the equation of the ellipse whose vertices
are at (6, 0) and foci at (4, 0).
(2, 0) and the eccentricity is
Find the equation of the hyperbola whose foci
are (
Q.29
Q.31
Find the equation of the hyperbola with vertices
Q.48
Find the equation of the hyperbola having its
foci at (0,
point P(3, 4).
14 ) and passing through the Page # 14
CONIC SECTIONS
ANSWERS
1. x2+y24x8y5 = 0
2. Centre (3, 1), radius = 3
29.
x2 y2
1
36 20
30.
x2 y2
1
16 9
31.
x2 y2
1
16 12
32.
x2 y2
1
9 25
33. 7x2 + 15y2 = 247
34.
x2 y2
1
18 9
3. x2 + y2 4x + 10y 21 = 0
4. (x2 + y2 12y + 11 = 0) or (x2 + y2 + 4y 21 = 0)
5. x +y –4x+6y+3 = 0
2
6. x +y –4x+2y–20 = 0
2
2
2
7. 4x2+4y2–88x–16y+451=0
35.
8. x2 + y2 5x 7y + 16 = 0
4 x2 4 y 2
1
81
45
(ii) A(3, 0), B(3, 0)
36. (i) 6 units, 8 units
(iii) F1(5, 0), F2(5, 0)
9. x +y –2x–3y–18 = 0
2
2
10. Centre (2, 3), radius
(iv) e =
3 2
5
3
(v)
10
2
units
3
12. x2+y2–2x+2y–23 = 0, centre (1,–1) and radius=5
x2 y2
1
37.
36 28
13. x2 + y2 + 4x + 6y 85 = 0
y 2 x2
1
38.
25 39
14. x2+y2–10x+6y+9=0, centre C(5,–3) and radius=5
15. x2 + y2 –3x –2y –1 = 0
16. x + y 4x 6y + 9 = 0
2
2
39.
x2 y2
1
25 4
40.
x2 y 2
5
1, e
9 16
3
41.
x2 y 2
1
25 20
42.
x2 y2
1
4 12
43.
x2 y2
1
3
2
44. 16x2 2y2 = 1
17. x2 + y2 – 6x + 12y – 15 = 0
19. x2 + y2 ax by = 0
20. x2 + y2 17x 19y + 50 = 0
21. B(7, - 8)
22. (i) F(3, 0), O(0, 0), x + 3 = 0, y = 0, 12 units
(ii) F(2, 0), O(0, 0), x = 2, y = 0, 8 units
(iii) F(0, 4), O(0, 0), y + 4 = 0, x = 0, 16 units
(iv) F(0, 2), O(0, 0), y = 2, x = 0, 8 units
23. y2 = 8x 24. y2=16x
25. x2= 20y
26. 2x2=25y
27. 3x2=–4y
28. (i) 10 units, 6 units (ii) A(5, 0) and B(5, 0)
(iii) F1(4, 0) and F2(4, 0) (iv) e =
4
(v) 3.6 units
5
45. x –y =32
2
47.
2
y2 x2
1
64 36
y 2 x2
1
46.
25 144
48. y2 x2 = 7
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