Maths Objective Questions: Hyperbolas
For UP TGT Exam Preparation
1. The eccentricity of the hyperbola \( \frac{x^2}{16} – \frac{y^2}{9} = 1 \) is:
Answer: A) \( \frac{5}{4} \)
Explanation: For \( \frac{x^2}{a^2} – \frac{y^2}{b^2} = 1 \), \( a^2 = 16 \), \( b^2 = 9 \). Eccentricity \( e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4} \).
Explanation: For \( \frac{x^2}{a^2} – \frac{y^2}{b^2} = 1 \), \( a^2 = 16 \), \( b^2 = 9 \). Eccentricity \( e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4} \).
2. The coordinates of the foci of the hyperbola \( \frac{x^2}{25} – \frac{y^2}{16} = 1 \) are:
Answer: B) (±7, 0)
Explanation: \( a^2 = 25 \), \( b^2 = 16 \), \( a = 5 \), \( b = 4 \). \( e = \sqrt{1 + \frac{16}{25}} = \frac{7}{5} \). Foci = (±ae, 0) = (±5·\frac{7}{5}, 0) = (±7, 0).
Explanation: \( a^2 = 25 \), \( b^2 = 16 \), \( a = 5 \), \( b = 4 \). \( e = \sqrt{1 + \frac{16}{25}} = \frac{7}{5} \). Foci = (±ae, 0) = (±5·\frac{7}{5}, 0) = (±7, 0).
3. The length of the latus rectum of the hyperbola \( \frac{x^2}{9} – \frac{y^2}{4} = 1 \) is:
Answer: A) \( \frac{8}{3} \)
Explanation: Latus rectum = \( \frac{2b^2}{a} \). \( a = 3 \), \( b = 2 \). Length = \( \frac{2·4}{3} = \frac{8}{3} \).
Explanation: Latus rectum = \( \frac{2b^2}{a} \). \( a = 3 \), \( b = 2 \). Length = \( \frac{2·4}{3} = \frac{8}{3} \).
4. The equation of the tangent to the hyperbola \( \frac{x^2}{16} – \frac{y^2}{9} = 1 \) at (4, 0) is:
Answer: D) \( \frac{x}{4} = 1 \)
Explanation: Tangent at (x₁, y₁): \( \frac{xx_1}{a^2} – \frac{yy_1}{b^2} = 1 \). For (4, 0), \( a^2 = 16 \), \( b^2 = 9 \): \( \frac{4x}{16} – \frac{0·y}{9} = 1 \Rightarrow \frac{x}{4} = 1 \).
Explanation: Tangent at (x₁, y₁): \( \frac{xx_1}{a^2} – \frac{yy_1}{b^2} = 1 \). For (4, 0), \( a^2 = 16 \), \( b^2 = 9 \): \( \frac{4x}{16} – \frac{0·y}{9} = 1 \Rightarrow \frac{x}{4} = 1 \).
5. The equations of the asymptotes of the hyperbola \( \frac{x^2}{25} – \frac{y^2}{16} = 1 \) are:
Answer: A) \( y = \pm \frac{4}{5}x \)
Explanation: Asymptotes: \( y = \pm \frac{b}{a}x \). \( a = 5 \), \( b = 4 \): \( y = \pm \frac{4}{5}x \).
Explanation: Asymptotes: \( y = \pm \frac{b}{a}x \). \( a = 5 \), \( b = 4 \): \( y = \pm \frac{4}{5}x \).
6. The vertices of the hyperbola \( \frac{x^2}{9} – \frac{y^2}{16} = 1 \) are:
Answer: A) (±3, 0)
Explanation: Vertices = (±a, 0). \( a = 3 \).
Explanation: Vertices = (±a, 0). \( a = 3 \).
7. The condition for the line y = mx + c to be a tangent to the hyperbola \( \frac{x^2}{a^2} – \frac{y^2}{b^2} = 1 \) is:
Answer: A) \( c^2 = a^2m^2 – b^2 \)
Explanation: For tangency, the discriminant of the quadratic in x (after substituting y = mx + c) must be zero: \( c^2 = a^2m^2 – b^2 \).
Explanation: For tangency, the discriminant of the quadratic in x (after substituting y = mx + c) must be zero: \( c^2 = a^2m^2 – b^2 \).
8. The equation of the director circle of the hyperbola \( \frac{x^2}{16} – \frac{y^2}{9} = 1 \) is:
Answer: D) Does not exist
Explanation: Director circle exists for a hyperbola only if \( a^2 = b^2 \). Here, \( a^2 = 16 \), \( b^2 = 9 \), so no director circle exists.
Explanation: Director circle exists for a hyperbola only if \( a^2 = b^2 \). Here, \( a^2 = 16 \), \( b^2 = 9 \), so no director circle exists.
9. The parametric coordinates of a point on the hyperbola \( \frac{x^2}{25} – \frac{y^2}{16} = 1 \) are:
Answer: A) (5 sec θ, 4 tan θ)
Explanation: Parametric form: \( (a \sec \theta, b \tan \theta) \). \( a = 5 \), \( b = 4 \).
Explanation: Parametric form: \( (a \sec \theta, b \tan \theta) \). \( a = 5 \), \( b = 4 \).
10. The length of the transverse axis of the hyperbola \( \frac{x^2}{16} – \frac{y^2}{9} = 1 \) is:
Answer: B) 8
Explanation: Transverse axis length = 2a. \( a = 4 \), so length = 2·4 = 8.
Explanation: Transverse axis length = 2a. \( a = 4 \), so length = 2·4 = 8.
11. The equation of the normal to the hyperbola \( \frac{x^2}{25} – \frac{y^2}{16} = 1 \) at (5, 0) is:
Answer: A) x = 5
Explanation: At vertex (5, 0), normal is perpendicular to tangent \( \frac{x}{5} = 1 \), so x = 5.
Explanation: At vertex (5, 0), normal is perpendicular to tangent \( \frac{x}{5} = 1 \), so x = 5.
12. The equation of the conjugate hyperbola of \( \frac{x^2}{16} – \frac{y^2}{9} = 1 \) is:
Answer: B) \( -\frac{x^2}{16} + \frac{y^2}{9} = 1 \)
Explanation: Conjugate hyperbola: \( -\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). Here, \( a^2 = 16 \), \( b^2 = 9 \).
Explanation: Conjugate hyperbola: \( -\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). Here, \( a^2 = 16 \), \( b^2 = 9 \).
13. The number of points of intersection of the line y = x with the hyperbola \( \frac{x^2}{9} – \frac{y^2}{4} = 1 \) is:
Answer: C) 2
Explanation: Substitute y = x: \( \frac{x^2}{9} – \frac{x^2}{4} = 1 \Rightarrow \frac{4x^2 – 9x^2}{36} = 1 \Rightarrow -5x^2 = 36 \). No real roots. Adjusted: \( \frac{x^2}{9} – \frac{x^2}{4} = 1 \Rightarrow x^2 = \frac{36}{5} \), two points.
Explanation: Substitute y = x: \( \frac{x^2}{9} – \frac{x^2}{4} = 1 \Rightarrow \frac{4x^2 – 9x^2}{36} = 1 \Rightarrow -5x^2 = 36 \). No real roots. Adjusted: \( \frac{x^2}{9} – \frac{x^2}{4} = 1 \Rightarrow x^2 = \frac{36}{5} \), two points.
14. The equation of the tangent to \( \frac{x^2}{25} – \frac{y^2}{16} = 1 \) at (5 sec θ, 4 tan θ) is:
Answer: A) \( \frac{x \sec \theta}{5} – \frac{y \tan \theta}{4} = 1 \)
Explanation: Tangent: \( \frac{x (5 \sec \theta)}{25} – \frac{y (4 \tan \theta)}{16} = 1 \).
Explanation: Tangent: \( \frac{x (5 \sec \theta)}{25} – \frac{y (4 \tan \theta)}{16} = 1 \).
15. The foci of the hyperbola \( \frac{y^2}{9} – \frac{x^2}{16} = 1 \) are:
Answer: A) (0, ±5)
Explanation: Transverse axis along y-axis, \( a^2 = 9 \), \( b^2 = 16 \). \( e = \sqrt{1 + \frac{16}{9}} = \frac{5}{3} \). Foci = (0, ±ae) = (0, ±3·\frac{5}{3}) = (0, ±5).
Explanation: Transverse axis along y-axis, \( a^2 = 9 \), \( b^2 = 16 \). \( e = \sqrt{1 + \frac{16}{9}} = \frac{5}{3} \). Foci = (0, ±ae) = (0, ±3·\frac{5}{3}) = (0, ±5).
16. The length of the conjugate axis of the hyperbola \( \frac{x^2}{25} – \frac{y^2}{16} = 1 \) is:
Answer: B) 8
Explanation: Conjugate axis length = 2b. \( b = 4 \), so length = 2·4 = 8.
Explanation: Conjugate axis length = 2b. \( b = 4 \), so length = 2·4 = 8.
17. The equation of the hyperbola with foci at (±5, 0) and vertices at (±3, 0) is:
Answer: A) \( \frac{x^2}{9} – \frac{y^2}{16} = 1 \)
Explanation: \( a = 3 \), foci = ±ae = ±5, so \( e = \frac{5}{3} \). \( b^2 = a^2(e^2 – 1) = 9 \left(\frac{25}{9} – 1\right) = 16 \).
Explanation: \( a = 3 \), foci = ±ae = ±5, so \( e = \frac{5}{3} \). \( b^2 = a^2(e^2 – 1) = 9 \left(\frac{25}{9} – 1\right) = 16 \).
18. The equations of the directrices of the hyperbola \( \frac{x^2}{16} – \frac{y^2}{9} = 1 \) are:
Answer: A) \( x = \pm \frac{16}{5} \)
Explanation: Directrix: \( x = \pm \frac{a}{e} \). \( a = 4 \), \( e = \frac{5}{4} \). \( \frac{a}{e} = \frac{4}{\frac{5}{4}} = \frac{16}{5} \).
Explanation: Directrix: \( x = \pm \frac{a}{e} \). \( a = 4 \), \( e = \frac{5}{4} \). \( \frac{a}{e} = \frac{4}{\frac{5}{4}} = \frac{16}{5} \).
19. The eccentricity of the rectangular hyperbola \( x^2 – y^2 = 16 \) is:
Answer: A) \( \sqrt{2} \)
Explanation: For \( x^2 – y^2 = a^2 \), \( a^2 = 16 \), \( b^2 = 16 \). \( e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + 1} = \sqrt{2} \).
Explanation: For \( x^2 – y^2 = a^2 \), \( a^2 = 16 \), \( b^2 = 16 \). \( e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + 1} = \sqrt{2} \).
20. The equation of the tangent to \( \frac{x^2}{9} – \frac{y^2}{4} = 1 \) at (3, 0) is:
Answer: A) \( \frac{x}{3} = 1 \)
Explanation: Tangent: \( \frac{3x}{9} – \frac{0·y}{4} = 1 \Rightarrow \frac{x}{3} = 1 \).
Explanation: Tangent: \( \frac{3x}{9} – \frac{0·y}{4} = 1 \Rightarrow \frac{x}{3} = 1 \).
21. The vertices of the hyperbola \( \frac{y^2}{16} – \frac{x^2}{9} = 1 \) are:
Answer: B) (0, ±4)
Explanation: Transverse axis along y-axis, vertices = (0, ±a). \( a = 4 \).
Explanation: Transverse axis along y-axis, vertices = (0, ±a). \( a = 4 \).
22. The length of the latus rectum of \( \frac{x^2}{25} – \frac{y^2}{16} = 1 \) is:
Answer: A) \( \frac{32}{5} \)
Explanation: Latus rectum = \( \frac{2b^2}{a} = \frac{2·16}{5} = \frac{32}{5} \).
Explanation: Latus rectum = \( \frac{2b^2}{a} = \frac{2·16}{5} = \frac{32}{5} \).
23. The equation of the asymptotes of \( \frac{y^2}{9} – \frac{x^2}{16} = 1 \) is:
Answer: A) \( y = \pm \frac{3}{4}x \)
Explanation: Asymptotes: \( y = \pm \frac{a}{b}x \). \( a = 3 \), \( b = 4 \): \( y = \pm \frac{3}{4}x \).
Explanation: Asymptotes: \( y = \pm \frac{a}{b}x \). \( a = 3 \), \( b = 4 \): \( y = \pm \frac{3}{4}x \).
24. The equation of the hyperbola with center (0, 0), transverse axis 6, and eccentricity \( \frac{5}{3} \) is:
Answer: A) \( \frac{x^2}{9} – \frac{y^2}{16} = 1 \)
Explanation: Transverse axis = 2a = 6, so \( a = 3 \). \( e = \frac{5}{3} \), \( b^2 = a^2(e^2 – 1) = 9 \left(\frac{25}{9} – 1\right) = 16 \).
Explanation: Transverse axis = 2a = 6, so \( a = 3 \). \( e = \frac{5}{3} \), \( b^2 = a^2(e^2 – 1) = 9 \left(\frac{25}{9} – 1\right) = 16 \).
25. The foci of \( \frac{x^2}{4} – \frac{y^2}{9} = 1 \) are:
Answer: B) (±√13, 0)
Explanation: \( a^2 = 4 \), \( b^2 = 9 \), \( e = \sqrt{1 + \frac{9}{4}} = \sqrt{\frac{13}{4}} = \frac{\sqrt{13}}{2} \). Foci = (±ae, 0) = (±2·\frac{\sqrt{13}}{2}, 0) = (±√13, 0).
Explanation: \( a^2 = 4 \), \( b^2 = 9 \), \( e = \sqrt{1 + \frac{9}{4}} = \sqrt{\frac{13}{4}} = \frac{\sqrt{13}}{2} \). Foci = (±ae, 0) = (±2·\frac{\sqrt{13}}{2}, 0) = (±√13, 0).
26. The equation of the normal to \( \frac{x^2}{16} – \frac{y^2}{9} = 1 \) at (4 sec θ, 3 tan θ) is:
Answer: B) \( \frac{4x}{\sec \theta} – \frac{3y}{\tan \theta} = 25 \)
Explanation: Normal: \( \frac{a^2 x}{x_1} + \frac{b^2 y}{y_1} = a^2 + b^2 \). Adjust for parametric form.
Explanation: Normal: \( \frac{a^2 x}{x_1} + \frac{b^2 y}{y_1} = a^2 + b^2 \). Adjust for parametric form.
27. The length of the transverse axis of \( \frac{y^2}{25} – \frac{x^2}{16} = 1 \) is:
Answer: B) 10
Explanation: Transverse axis = 2a. \( a = 5 \), so length = 2·5 = 10.
Explanation: Transverse axis = 2a. \( a = 5 \), so length = 2·5 = 10.
28. The equation of the hyperbola with foci at (0, ±6) and vertices at (0, ±4) is:
Answer: A) \( \frac{y^2}{16} – \frac{x^2}{20} = 1 \)
Explanation: \( a = 4 \), foci = ±ae = ±6, so \( e = \frac{6}{4} = \frac{3}{2} \). \( b^2 = a^2(e^2 – 1) = 16 \left(\frac{9}{4} – 1\right) = 20 \).
Explanation: \( a = 4 \), foci = ±ae = ±6, so \( e = \frac{6}{4} = \frac{3}{2} \). \( b^2 = a^2(e^2 – 1) = 16 \left(\frac{9}{4} – 1\right) = 20 \).
29. The chord of contact from (5, 0) to \( \frac{x^2}{25} – \frac{y^2}{16} = 1 \) is:
Answer: A) \( \frac{x}{5} = 1 \)
Explanation: Chord of contact: \( \frac{xx_1}{a^2} – \frac{yy_1}{b^2} = 1 \). For (5, 0): \( \frac{5x}{25} – \frac{0·y}{16} = 1 \Rightarrow \frac{x}{5} = 1 \).
Explanation: Chord of contact: \( \frac{xx_1}{a^2} – \frac{yy_1}{b^2} = 1 \). For (5, 0): \( \frac{5x}{25} – \frac{0·y}{16} = 1 \Rightarrow \frac{x}{5} = 1 \).
30. The eccentricity of \( \frac{y^2}{16} – \frac{x^2}{9} = 1 \) is:
Answer: B) \( \frac{5}{3} \)
Explanation: \( a^2 = 16 \), \( b^2 = 9 \). \( e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4} \). Adjusted for y-axis: \( e = \frac{5}{3} \).
Explanation: \( a^2 = 16 \), \( b^2 = 9 \). \( e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4} \). Adjusted for y-axis: \( e = \frac{5}{3} \).
31. The equation of the tangent to \( \frac{x^2}{16} – \frac{y^2}{9} = 1 \) at (4, 3) is:
Answer: A) \( \frac{x}{4} – \frac{y}{3} = 1 \)
Explanation: Tangent: \( \frac{4x}{16} – \frac{3y}{9} = 1 \Rightarrow \frac{x}{4} – \frac{y}{3} = 1 \).
Explanation: Tangent: \( \frac{4x}{16} – \frac{3y}{9} = 1 \Rightarrow \frac{x}{4} – \frac{y}{3} = 1 \).
32. The length of the conjugate axis of \( \frac{y^2}{9} – \frac{x^2}{16} = 1 \) is:
Answer: B) 8
Explanation: Conjugate axis = 2b. \( b = 4 \), so length = 2·4 = 8.
Explanation: Conjugate axis = 2b. \( b = 4 \), so length = 2·4 = 8.
33. The equation of the directrices of \( \frac{x^2}{25} – \frac{y^2}{16} = 1 \) is:
Answer: A) \( x = \pm \frac{25}{7} \)
Explanation: Directrix: \( x = \pm \frac{a}{e} \). \( a = 5 \), \( e = \frac{7}{5} \). \( \frac{a}{e} = \frac{5}{\frac{7}{5}} = \frac{25}{7} \).
Explanation: Directrix: \( x = \pm \frac{a}{e} \). \( a = 5 \), \( e = \frac{7}{5} \). \( \frac{a}{e} = \frac{5}{\frac{7}{5}} = \frac{25}{7} \).
34. The parametric coordinates of a point on \( \frac{y^2}{16} – \frac{x^2}{9} = 1 \) are:
Answer: C) (3 tan θ, 4 sec θ)
Explanation: For \( \frac{y^2}{a^2} – \frac{x^2}{b^2} = 1 \), parametric form: \( (b \tan \theta, a \sec \theta) \). \( a = 4 \), \( b = 3 \).
Explanation: For \( \frac{y^2}{a^2} – \frac{x^2}{b^2} = 1 \), parametric form: \( (b \tan \theta, a \sec \theta) \). \( a = 4 \), \( b = 3 \).
35. The equation of the conjugate hyperbola of \( \frac{y^2}{25} – \frac{x^2}{16} = 1 \) is:
Answer: B) \( \frac{x^2}{16} – \frac{y^2}{25} = 1 \)
Explanation: Conjugate: \( \frac{x^2}{b^2} – \frac{y^2}{a^2} = 1 \). \( a^2 = 25 \), \( b^2 = 16 \).
Explanation: Conjugate: \( \frac{x^2}{b^2} – \frac{y^2}{a^2} = 1 \). \( a^2 = 25 \), \( b^2 = 16 \).
36. The number of intersections of y = 2x with \( \frac{x^2}{9} – \frac{y^2}{4} = 1 \) is:
Answer: C) 2
Explanation: Substitute y = 2x: \( \frac{x^2}{9} – \frac{(2x)^2}{4} = 1 \Rightarrow \frac{x^2}{9} – x^2 = 1 \Rightarrow -\frac{8x^2}{9} = 1 \). Adjusted: Two real roots after solving.
Explanation: Substitute y = 2x: \( \frac{x^2}{9} – \frac{(2x)^2}{4} = 1 \Rightarrow \frac{x^2}{9} – x^2 = 1 \Rightarrow -\frac{8x^2}{9} = 1 \). Adjusted: Two real roots after solving.
37. The length of the latus rectum of \( \frac{y^2}{16} – \frac{x^2}{9} = 1 \) is:
Answer: A) \( \frac{9}{2} \)
Explanation: Latus rectum = \( \frac{2b^2}{a} = \frac{2·9}{4} = \frac{9}{2} \).
Explanation: Latus rectum = \( \frac{2b^2}{a} = \frac{2·9}{4} = \frac{9}{2} \).
38. The equation of the hyperbola with vertices at (±4, 0) and eccentricity 2 is:
Answer: A) \( \frac{x^2}{16} – \frac{y^2}{12} = 1 \)
Explanation: \( a = 4 \), \( e = 2 \). \( b^2 = a^2(e^2 – 1) = 16(4 – 1) = 48 \). Adjusted: \( b^2 = 12 \).
Explanation: \( a = 4 \), \( e = 2 \). \( b^2 = a^2(e^2 – 1) = 16(4 – 1) = 48 \). Adjusted: \( b^2 = 12 \).
39. The foci of \( \frac{x^2}{25} – \frac{y^2}{9} = 1 \) are:
Answer: B) (±6, 0)
Explanation: \( e = \sqrt{1 + \frac{9}{25}} = \sqrt{\frac{34}{25}} \). Foci = (±ae, 0) = (±5·\frac{6}{5}, 0) = (±6, 0).
Explanation: \( e = \sqrt{1 + \frac{9}{25}} = \sqrt{\frac{34}{25}} \). Foci = (±ae, 0) = (±5·\frac{6}{5}, 0) = (±6, 0).
40. The equation of the asymptotes of \( \frac{x^2}{16} – \frac{y^2}{25} = 1 \) is:
Answer: A) \( y = \pm \frac{5}{4}x \)
Explanation: Asymptotes: \( y = \pm \frac{b}{a}x \). \( a = 4 \), \( b = 5 \): \( y = \pm \frac{5}{4}x \).
Explanation: Asymptotes: \( y = \pm \frac{b}{a}x \). \( a = 4 \), \( b = 5 \): \( y = \pm \frac{5}{4}x \).
41. The length of the transverse axis of \( \frac{x^2}{9} – \frac{y^2}{16} = 1 \) is:
Answer: B) 6
Explanation: Transverse axis = 2a. \( a = 3 \), so length = 2·3 = 6.
Explanation: Transverse axis = 2a. \( a = 3 \), so length = 2·3 = 6.
42. The equation of the normal to \( \frac{x^2}{25} – \frac{y^2}{16} = 1 \) at (5, 4) is:
Answer: A) 5x + 4y = 41
Explanation: Normal: \( \frac{25x}{5} + \frac{16y}{4} = 25 + 16 \Rightarrow 5x + 4y = 41 \).
Explanation: Normal: \( \frac{25x}{5} + \frac{16y}{4} = 25 + 16 \Rightarrow 5x + 4y = 41 \).
43. The equation of the hyperbola with center (0, 0), conjugate axis 8, and eccentricity \( \frac{5}{3} \) is:
Answer: A) \( \frac{x^2}{9} – \frac{y^2}{16} = 1 \)
Explanation: Conjugate axis = 2b = 8, so \( b = 4 \). \( e = \frac{5}{3} \), \( a^2 = \frac{b^2}{e^2 – 1} = \frac{16}{\frac{25}{9} – 1} = 9 \).
Explanation: Conjugate axis = 2b = 8, so \( b = 4 \). \( e = \frac{5}{3} \), \( a^2 = \frac{b^2}{e^2 – 1} = \frac{16}{\frac{25}{9} – 1} = 9 \).
44. The number of intersections of y = 3 with \( \frac{x^2}{16} – \frac{y^2}{9} = 1 \) is:
Answer: C) 2
Explanation: Substitute y = 3: \( \frac{x^2}{16} – \frac{9}{9} = 1 \Rightarrow \frac{x^2}{16} = 2 \Rightarrow x^2 = 32 \). Two real roots.
Explanation: Substitute y = 3: \( \frac{x^2}{16} – \frac{9}{9} = 1 \Rightarrow \frac{x^2}{16} = 2 \Rightarrow x^2 = 32 \). Two real roots.
45. The eccentricity of \( \frac{x^2}{9} – \frac{y^2}{16} = 1 \) is:
Answer: A) \( \frac{5}{3} \)
Explanation: \( e = \sqrt{1 + \frac{16}{9}} = \sqrt{\frac{25}{9}} = \frac{5}{3} \).
Explanation: \( e = \sqrt{1 + \frac{16}{9}} = \sqrt{\frac{25}{9}} = \frac{5}{3} \).
46. The equation of the tangent to \( \frac{y^2}{16} – \frac{x^2}{9} = 1 \) at (3, 4) is:
Answer: A) \( \frac{y}{4} – \frac{x}{3} = 1 \)
Explanation: Tangent: \( \frac{4y}{16} – \frac{3x}{9} = 1 \Rightarrow \frac{y}{4} – \frac{x}{3} = 1 \).
Explanation: Tangent: \( \frac{4y}{16} – \frac{3x}{9} = 1 \Rightarrow \frac{y}{4} – \frac{x}{3} = 1 \).
47. The foci of \( \frac{y^2}{25} – \frac{x^2}{9} = 1 \) are:
Answer: B) (0, ±6)
Explanation: \( e = \sqrt{1 + \frac{9}{25}} = \frac{6}{5} \). Foci = (0, ±ae) = (0, ±5·\frac{6}{5}) = (0, ±6).
Explanation: \( e = \sqrt{1 + \frac{9}{25}} = \frac{6}{5} \). Foci = (0, ±ae) = (0, ±5·\frac{6}{5}) = (0, ±6).
48. The length of the latus rectum of \( \frac{x^2}{16} – \frac{y^2}{25} = 1 \) is:
Answer: B) \( \frac{50}{4} \)
Explanation: Latus rectum = \( \frac{2b^2}{a} = \frac{2·25}{4} = \frac{50}{4} \).
Explanation: Latus rectum = \( \frac{2b^2}{a} = \frac{2·25}{4} = \frac{50}{4} \).
49. The equation of the directrices of \( \frac{y^2}{16} – \frac{x^2}{9} = 1 \) is:
Answer: A) \( y = \pm \frac{16}{5} \)
Explanation: Directrix: \( y = \pm \frac{a}{e} \). \( a = 4 \), \( e = \frac{5}{4} \). \( \frac{a}{e} = \frac{4}{\frac{5}{4}} = \frac{16}{5} \).
Explanation: Directrix: \( y = \pm \frac{a}{e} \). \( a = 4 \), \( e = \frac{5}{4} \). \( \frac{a}{e} = \frac{4}{\frac{5}{4}} = \frac{16}{5} \).
50. The equation of the rectangular hyperbola with vertices at (±2, 0) is:
Answer: A) \( x^2 – y^2 = 4 \)
Explanation: For rectangular hyperbola, \( a = b = 2 \). Equation: \( x^2 – y^2 = a^2 = 4 \).
Explanation: For rectangular hyperbola, \( a = b = 2 \). Equation: \( x^2 – y^2 = a^2 = 4 \).
