More practice questions

For Q5 to Q8, find the foci, the vertices, the eccentricity and the latus rectum.

5. Q5. 9y² − 4x² = 36 (NCERT Exercise 10.4, Q3).

   Show answerHide answer

   y²/4 − x²/9 = 1: a = 2, b = 3, c = √13. Foci (0, ±√13), vertices (0, ±2), e = √13/2, latus rectum 9.

6. Q6. 16x² − 9y² = 576 (NCERT Exercise 10.4, Q4).

   Show answerHide answer

   x²/36 − y²/64 = 1: a = 6, b = 8, c = 10. Foci (±10, 0), vertices (±6, 0), e = 5/3, latus rectum 64/3.

7. Q7. 5y² − 9x² = 36 (NCERT Exercise 10.4, Q5).

   Show answerHide answer

   y²/(36/5) − x²/4 = 1: a = 6/√5, b = 2, c² = 36/5 + 4 = 56/5. Foci (0, ±√(56/5)), vertices (0, ±6/√5), e = √14/3, latus rectum 2 × 4/(6/√5) = 4√5/3.

8. Q8. 49y² − 16x² = 784 (NCERT Exercise 10.4, Q6).

   Show answerHide answer

   y²/16 − x²/49 = 1: a = 4, b = 7, c = √65. Foci (0, ±√65), vertices (0, ±4), e = √65/4, latus rectum 49/2.

9. Q9. Find the hyperbola with vertices (0, ±5) and foci (0, ±8) (NCERT Exercise 10.4, Q8).

   Show answerHide answer

   b² = 64 − 25 = 39: y²/25 − x²/39 = 1.

10. Q10. Find the hyperbola with vertices (0, ±3) and foci (0, ±5) (NCERT Exercise 10.4, Q9).

    Show answerHide answer

    b² = 25 − 9 = 16: y²/9 − x²/16 = 1.

11. Q11. Find the hyperbola with foci (0, ±13) and conjugate axis of length 24 (NCERT Exercise 10.4, Q11).

    Show answerHide answer

    b = 12, a² = 169 − 144 = 25: y²/25 − x²/144 = 1.

12. Q12. Find the hyperbola with foci (±3√5, 0) and latus rectum of length 8 (NCERT Exercise 10.4, Q12).

    Show answerHide answer

    c² = 45 and 2b²/a = 8, so b² = 4a. Then a² + 4a − 45 = 0, (a + 9)(a − 5) = 0, a = 5 and b² = 20:

    x²/25 − y²/20 = 1.

13. Q13. Find the hyperbola with foci (±4, 0) and latus rectum of length 12 (NCERT Exercise 10.4, Q13).

    Show answerHide answer

    b² = 6a and a² + 6a = 16, so (a + 8)(a − 2) = 0, a = 2 and b² = 12:

    x²/4 − y²/12 = 1.

14. Q14. Find the eccentricity of an equilateral hyperbola, and the angle between its asymptotes.

    Show answerHide answer

    a = b gives c² = 2a², so e = √2. The asymptotes y = ±x are perpendicular: the angle is 90°.