JEE MAIN - Mathematics (2022 - 28th July Evening Shift - No. 12)
Let the hyperbola $$H: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$$ pass through the point $$(2 \sqrt{2},-2 \sqrt{2})$$. A parabola is drawn whose focus is same as the focus of $$\mathrm{H}$$ with positive abscissa and the directrix of the parabola passes through the other focus of $$\mathrm{H}$$. If the length of the latus rectum of the parabola is e times the length of the latus rectum of $$\mathrm{H}$$, where e is the eccentricity of H, then which of the following points lies on the parabola?
$$(2 \sqrt{3}, 3 \sqrt{2})$$
$$\mathbf(3 \sqrt{3},-6 \sqrt{2})$$
$$(\sqrt{3},-\sqrt{6})$$
$$(3 \sqrt{6}, 6 \sqrt{2})$$
Explanation
$$H:{{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$
Focus of parabola : $$(ae,\,0)$$
Directrix : $$x = - ae$$.
Equation of parabola $$ \equiv {y^2} = 4aex$$
Length of latus rectum of parabola $$ = 4ae$$
Length of latus rectum of hyperbola $$ = {{2.{b^2}} \over a}$$
as given, $$4ae = {{2{b^2}} \over a}\,.\,e$$
$$2 = {{{b^2}} \over {{a^2}}}$$ ...... (i)
$$\because$$ H passes through $$\left( {2\sqrt 2 , - 2\sqrt 2 } \right) \Rightarrow {8 \over {{a^2}}} - {8 \over {{b^2}}} = 1$$ ........ (ii)
From (i) and (ii) $${a^2} = 4$$ and $${b^2} = 8 \Rightarrow e = \sqrt 3 $$
$$\Rightarrow$$ Equation of parabola is $${y^2} = 8\sqrt 3 x$$.
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