Question

If C is the centre and A, B are two points on the conic 4x2 + 9y2 – 8x – 36y + 4 = 0 such that ŠACB = $\frac{\pi}{2}$, 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 centred 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 (4cos2q + 9 sin2q) = 36 and

CB2 (4sin2q + 9cos2q) = 36 giving

CA–2 + CB–2=$\frac { 13 } { 36 }$