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A normal to the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ meets the transverse and conjugate axes in M and N and the lines MP and NP are drawn at right angles to the axes. The locus of P is

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the ellipse $b^{2} x^{2} + a^{2} y^{2} = a^{2} + b^{2}$

the hyperbola $a^{2} x^{2} - b^{2} y^{2} = {(a^{2} + b^{2})}^{2}$

the circle $x^{2} + y^{2} = ab$

the parabola $y^{2} = 4 a (x + b)$

answer is D.

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Detailed Solution

Let $P (x_{1} y_{1}) \Rightarrow M (x_{1}, 0), N (0, y_{1})$ ;

Equation of line $\bar{MN} is, \frac{x}{x_{1}} + \frac{y}{y_{1}} = 1 . . . . . . . . . . (1)$

(1) is normal $\Rightarrow \frac{a^{2}}{l^{2}} - \frac{b^{2}}{m^{2}} = \frac{{(a^{2} + b^{2})}^{2}}{n^{2}} \Rightarrow \frac{a^{2}}{\frac{1}{{x^{2}}_{1}}} - \frac{b^{2}}{\frac{1}{{y^{2}}_{1}}} = \frac{{(a^{2} + b^{2})}^{2}}{{(- 1)}^{2}}$

locus is $a^{2} x^{2} - b^{2} y^{2} = {(a^{2} + b^{2})}^{2}$

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