A tangent at any point P on the rectangular hyperbola $x y = c^{2}$ meets the lines y = x and y = - x at Q and R respectively. The normal at P meets X and Y axes at M and N respectively. If $A_{1}, A_{2}$ be the areas of triangles OQR and OMN respectively, then which of the following is/are not independent of P?

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$A_{1} A_{2}$

$A_{1} A_{2}^{2}$

$A_{1} + A_{2}^{}$

$A_{1}^{2} A_{2}$

answer is A, C, D.

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Detailed Solution

$\begin{array}{l} P = (c t, \frac{c}{t}) tangent is x + y t^{2} = 2 c t \to (1) \\ Normal is t^{3} x - t y = c (t^{4} - 1) \to (2) \\ s o l v e (1) w i t h t h e l i n e y = x, w e g e t Q a n d w i t h t h e l i n e y = - x, w e g e t R \\ Q = (\frac{2 c t}{1 + t^{2}}, \frac{2 c t}{1 + t^{2}}), R = (\frac{2 c t}{1 - t^{2}}, \frac{- 2 c t}{1 - t^{2}},) \end{array}$ ;

$n o w s o l v e (2) w i t h t h e l i n e y = 0, w e g e t R a n d w i t h t h e l i n e x = 0, w e g e t N$

$M = (\frac{c (t^{4} - 1)}{t^{3}}, 0)$ $N = (0, \frac{c (1 - t^{4})}{t})$

$A r e a o f t r i a n g l e O Q R = A_{1} = \frac{4 c^{2} t^{2}}{(1 - t^{4})}, A r e a o f t r i a n g l e O M N = A_{2} = \frac{c^{2} {(1 - t^{4})}^{2}}{2 t^{4}},$ $(\sin c e a r e a o f t r i a n g l e f o r m e d b y t h e v e r t i c e s (0, 0), (x_{1,} y_{1}), (x_{2,} y_{2}) i s = \frac{1}{2} |x_{1} y_{2} - x_{2} y_{1}|)$