Conic sections #5 - Questions and Answers

Question 1 How do I find the directrix and focus of a parabola? Answer There are two basic kinds of parabola (as it is convenient for me to say) Type 1: The parabola lying on, or parallel to the x-axis This parabola is of the form (y - yv)² = 4a(x – xv) Where, focus is (a + xv, yv) directrix is the line x = xv - a Vertex is (xv, yv) Type 2: The parabola lying on, or parallel to the y-axis This parabola is of the form (x - xv)² = 4a(y - yv) Where, focus is (xv, a + yv) directrix is the line y = yv - a Vertex is (xv, yv) Question 2 The polar equation of the conic with the given eccentricity and directrix and focus at origin: r = 4 / (1 + cos θ) Answer Step 1 If the conic of directrix x = p where p is positive real number and eccentricity is e, and focus at origin then the polar equation of conic is: r = ep / (1 ± e cos θ) Step 2 It is provided that eccentricity is e = 1 and directrix is x = 4 and focus at the origin. To find the polar equation of conic put the provided value in above equation as: r = 1 × 4 / (1 ± 1 × cos θ) Since, here directrix is positive x = 4 so, consider the positive sign as: r = 4 / (1 + cos θ) Therefore, polar equation is r = 4 / (1 + cos θ) Question 3 To find: The equation of the given hyperbola. The foci of the hyperbola is (0, ± 5) and eccentricity is e = 1.5. Answer Step 1 Write the general equation of the hyperbola. (y - k)² / a² - (x - h)² / b² = 1 The center of the hyperbola is (h, k), the vertices are (h ± a, k), and foci are (h ± c, k). Write the expression for c. c = √(a² + b²) Write the expression for the eccentricity of the hyperbola. e = √(a² + b²) / a. The foci of the conic section are, (h, k ± c) = (0, ± 5) h = 0 c = 5 k = 0 Step 2 Write the expression for the eccentricity of the hyperbola. e = c / a Substitute 5 for c and 1.5 for e. 1.5 = 5 / a a = 5 / 1.5 a = 3.33 Write the expression for c. c = √(a² + b²) Substitute 3.33 for a and 5 for c. 5 = √(3.33² + b²) 25 = 3.33² + b² 25 = 11.08 + b² b² = 13.92 b = 3.73 Write the general equation of the hyperbola. (y - k)² / a² - (x - h)² / b² = 1 Substitute 3.33 for a, 0 for h, 3.73 for b, and 0 for k. (y - 0)² / 3.33² - (x - 0)² / 3.73² = 1 y² / 11.08 - x² / 13.92 = 1 Therefore, the equation of the given hyperbola is y² / 11.08 - x² / 13.92 = 1. Question 4 To calculate the vertices and foci of the conic section: (x/9)² + (y/4)² = 1 Answer Step 1 Formula: eccentricity (e) = √(1 - (b²/a²)) Step 2 Calculation: Compare this equation with the standard ellipse equation x²/a² + y²/b² = 1 and we get: a² = 81 a = 9 b² = 16 ⇒ b = 4 Now, e = √(1 - b²/a²) = √(1 - 16/81) = √65/9 Vertices are: (± a, 0) and (0, ± b) Now, put the values of a and b to get the vertices of the conic section and we get: (± a, 0) and (0, ± b) = (± 9, 0) and (0, ± 4) Now, Foci = (± ae, 0) = (± 9(√65/9), 0) = (±√65, 0) Thus, the vertices and foci of the conic section are: (±9, 0) and (0, ± 4) and (± √65, 0). Question 5 We need to verify theorem 5 Focus-Directrix definition: 0 < e < 1 and c = d(e² - 1) Answer Step 1 A conic section is the set of all points in a plane whose distances from a particular point (focus) and a particular line (directrix) in the plane have a constant ratio. If P is a point of a conic section, F is the conic's focus and D is the point of the directrix closest to P, then e = PF/PD and PF = e × PD, where e is a constant and the eccentricity of the conic. Step 2 Formula: Assume P(x, y) is a point on the ellipse and PF = √((x - c)² + y²) so the equation PF = ePD will be written as. Step 3 Calculation: If is given that, c = d(e² - 1) and we know that a = ce. Assume the coordinates of the focus F = (c, 0) with vertical directrix x = ae, then x = d + c = c / (e² - 1) + c = c(e² / (e² - 1)) = (ae)/e = ae = x √((x - c)² + y²) = e(x - ae) (x - c)² + y² = e²(x - ae)² x² - 2cx + c² + y² = e²(x² - 2aex + a²e²) x² - 2cx + c² + y² = e²x² - 2ae³x + a²e^4 = e²x² - 2aex + a²(1 - e²) x²(1 - e²) + y² = - 2ae²x + 2cx + a²(1 - e²) x²(1 - e²) + y² = - 2aex + a² - a²e² + 2aex x²(1 - e²) + y² = a²(1 - e²) (1 - e²)x² / a²(1 - e²) + y² / a²(1 - e²) = 1 x²/a² + y²/a²(1 - e²) = 1 (x² / a²) + (y² / a²(1 - (c²/a²)) (x² / a²) + (y² / (a² - c²)) (x² / a²) + (y² / b²) = 1 Thus, the theorem 5 Focus-Directrix definition is verified.