Question

If c is the centre and A, B are two points on the conic 4x² + 9y² –8x –36y + 4 = 0 such that A$\widehat{C}$B =$\frac{\pi}{2}$, then CA^–2 + CB^–2 is equal to-

• A. $\frac{13}{36}$
• B. $\frac{36}{13}$
• C. $\frac{16}{33}$
• D. $\frac{33}{16}$

Answer 1

Answer: (A)

$\frac{13}{36}$

Answer 2

The equation can be written as 4(x –1)² + 9(y –2)² = 36 which is an ellipse centered at

(1, 2). If CA makes an angle q with the major axis, then

A ŗ [1 + CA cos q, 2 + CA sin q]

B ŗ $\left\lbrack 1 + CB\cos\left( \frac{\pi}{2} + \theta \right),2 + CB\sin\left( \frac{\pi}{2} + \theta \right) \right\rbrack$

As A and B are points on the conic CA² (4 cos² q + 9 sin² q) = 36 and

CB² (4 sin²q + 9 cos² q) = 36 giving

CA^–2 + CB^–2 = $\frac{13}{36}$