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?
Solution
<p>$H:{{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$</p>
<p>Focus of parabola : $(ae,\,0)$</p>
<p>Directrix : $x = - ae$.</p>
<p>Equation of parabola $\equiv {y^2} = 4aex$</p>
<p>Length of latus rectum of parabola $= 4ae$</p>
<p>Length of latus rectum of hyperbola $= {{2.{b^2}} \over a}$</p>
<p>as given, $4ae = {{2{b^2}} \over a}\,.\,e$</p>
<p>$2 = {{{b^2}} \over {{a^2}}}$ ...... (i)</p>
<p>$\because$ H passes through $$\left( {2\sqrt 2 , - 2\sqrt 2 } \right) \Rightarrow {8 \over {{a^2}}} - {8 \over {{b^2}}} = 1$$ ........ (ii)</p>
<p>From (i) and (ii) ${a^2} = 4$ and ${b^2} = 8 \Rightarrow e = \sqrt 3$</p>
<p>$\Rightarrow$ Equation of parabola is ${y^2} = 8\sqrt 3 x$.</p>
About this question
Subject: Mathematics · Chapter: Conic Sections · Topic: Parabola
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