Single Choice

The perpendicular from the centre upon the normal on any point of the hyperbola $$\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$$ meets at $$R$$ . Find the locus of $$R$$.

A$$(x^2\, +\, y^2)^2\, (a^2y^2\, -\, b^2x^2)\, =\, x^2y^2(a^2\, +\, b^2)^2$$
Correct Answer
B$$(x^2\, +\, y^2)^2\, (a^2y^2\, -\, b^2x^2)\, =\, x^2y^2(a^2\, -\, b^2)^2$$
C$$(x^2\, -\, y^2)^2\, (a^2y^2\, +\, b^2x^2)\, =\, x^2y^2(a^2\, -\, b^2)^2$$
D$$(x^2\, -\, y^2)^2\, (a^2y^2\, -\, b^2x^2)\, =\, x^2y^2(a^2\, +\, b^2)^2$$

Solution

Equation of normal to the given hyperbola is given by,
$$\displaystyle \frac{ax}{\sec\theta}+\frac{by}{\tan\theta}=a^2+b^2..(1)$$
Slope $$\displaystyle m = -\frac{a\tan\theta}{b\sec\theta}=-\frac{a\sin\theta}{b}$$
Thus equation of line through centre and perpendicular to the (1) is, $$y =-\cfrac{1}{m}x$$
$$\Rightarrow\displaystyle y = \frac{b}{a\sin\theta}x ..(2)$$
Thus by eliminating $$\theta$$ from (1) and (2) we will get required locus,
$$\Rightarrow \displaystyle \cos\theta\left(ax+\frac{by}{bx/ay}\right)=a^2+b^2$$
$$\Rightarrow \displaystyle a\cos\theta\left(x^2+y^2\right)=x(a^2+b^2)$$
Squaring we get,
$$\Rightarrow \displaystyle a^2\cos^2\theta\left(x^2+y^2\right)^2=x^2\left(a^2+b^2\right)^2$$
$$\Rightarrow \displaystyle a^2\left(1-\frac{b^2x^2}{a^2y^2}\right)\left(x^2+y^2\right)^2=x^2\left(a^2+b^2\right)^2$$
$$\Rightarrow (x^2\, +\, y^2)^2\, (a^2y^2\, -\, b^2x^2)\, =\, x^2y^2(a^2\, +\, b^2)^2$$

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