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  • Let the hyperbola  <img alt="\mathrm{H: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1}" src="/latex-image/?%5Cmathrm%7BH%3A%20%5Cfrac%7Bx%5E%7B2%7D%7D%7Ba%5E%7B2%7D%7D-%5Cfrac%7By%5E%7B2%7D%7

Let the hyperbola  \mathrm{H: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1} pass through the point \mathrm{(2 \sqrt{2},-2 \sqrt{2})}. A parabola is drawn whose focus is same as the focus of \mathrm{\mathrm{H}} with positive abscissa and the directrix of the parabola passes through the other focus of \mathrm{\mathrm{H}}. If the length of the latus rectum of the parabola is e times the length of the latus rectum of \mathrm{ \mathrm{H},} where e is the eccentricity of \mathrm{ \mathrm{H},} then which of the following points lies on the parabola?

Option: 1

(2 \sqrt{3}, 3 \sqrt{2})


Option: 2

(3 \sqrt{3},-6 \sqrt{2})


Option: 3

(\sqrt{3},-\sqrt{6})


Option: 4

(3 \sqrt{6}, 6 \sqrt{2})


Answers (1)

best_answer

\mathrm{(2 \sqrt{2},-2 \sqrt{2}) \Rightarrow \frac{8}{a^2}-\frac{b}{b^2}=1 \quad \cdots \text { (i) }}

 

\begin{aligned} & \text{Let Parabola be\: } \mathrm{y^2=4(a e) x}\\ \therefore \mathrm{4 a e} &=\mathrm{e \times \frac{2 b^2}{a}} \\ \mathrm{2 a^2 }&=\mathrm{b^2}\\ &\text{Put this in (i)}\\ & \mathrm{\frac{8}{a^2}-\frac{b}{2 a^2}=1} \\ \Rightarrow & \mathrm{\frac{4}{a^2}=1 \Rightarrow a^2=4 \Rightarrow b^2=8 }\\ \mathrm{e^2}=& \mathrm{1+\frac{b^2}{a^2}=3 \Rightarrow e=\sqrt{3}}\\ & \therefore \text{Parabola is\: }\mathrm{ y^2=8 \sqrt{3} x}\\ &\text{ option (B) satisfies it} \\ &\therefore \quad \text{option (B)} \end{aligned}

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Irshad Anwar

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