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Let the focal chord of the parabola \mathrm{\mathrm{P}: y^{2}=4 x} along the line \mathrm{\mathrm{L}: y=\mathrm{m} x+\mathrm{c}, \mathrm{m}>0} meet the parabola at the points \mathrm{\mathrm{M} \: and\: \mathrm{N}}. Let the line \mathrm{L} be a tangent to the hyperbola \mathrm{\mathrm{H}: x^{2}-y^{2}=4}. If \mathrm{O} is the vertex of \mathrm{\mathrm{P} \: and \: \mathrm{F}} is the focus of \mathrm{H} on the positive x-axis, then the area of the quadrilateral OMFN is :
 

Option: 1

2 \sqrt{6}


Option: 2

2 \sqrt{14}


Option: 3

4 \sqrt{6}


Option: 4

4 \sqrt{14}


Answers (1)

best_answer

\begin{aligned} &\mathrm{\quad \frac{x^{2}}{4}-\frac{y^{2}}{4}=1}\\ &\mathrm{forcus (a 0,0)}\\ &\mathrm{f(2 \sqrt{2}, 0)}\\ &\text{ line } \mathrm{L: y=m x+c }\text{pass (1,0)}\\ &\mathrm{0=m+c} \cdots \text { (1) }\\ &\text{Line L is tangent to Hyperbola}\mathrm{ \frac{x^{2}}{4}-\frac{y^{2}}{4}=1}\\ &\mathrm{c=\pm \sqrt{a^{2} m^{2}-l^{2}}} \\ &\mathrm{c=\pm \sqrt{4 m^{2}-4}}\\ &\text{from (1)}\\ \mathrm{-m }&\mathrm{=\pm \sqrt{4 m^{2}-4}} \\ \mathrm{m^{2}} &\mathrm{=4 m^{2}-4} \\ \mathrm{4} &\mathrm{=3 m^{2}} \\ \mathrm{\frac{2}{\sqrt{3}}} &\mathrm{=m }\quad(\text { as } \mathrm{m>0)} \\ \mathrm{c} &=\mathrm{-m} \\ \mathrm{c }&=\mathrm{-\frac{2}{\sqrt{3}}} \end{aligned}

\begin{aligned} & \mathrm{ y=\frac{2 x}{\sqrt{3}}-\frac{2}{\sqrt{3}}} \\ & \mathrm{y^{2}=4 x} \\ & \mathrm{\Rightarrow\left(\frac{2 x-2}{\sqrt{3}}\right)^{2}=4 x} \\ & \mathrm{\Rightarrow x^{2}+1-2 x=3 x }\\ & \mathrm{x^{2}-5 x+1=0 }\\ &\mathrm{ y^{2}=4\left(\frac{\sqrt{3} y+2}{2}\right)} \\ & \mathrm{y^{2}=2 \sqrt{3} y+4 }\\ & \mathrm{\Rightarrow y^{2}-2 \sqrt{3} y-4=0 }\\ & \text { Area }=\mathrm{\left|\frac{1}{2}\right| \begin{array}{ccccc}0 & x_{1} & 2 \sqrt{2} & x_{2} & 0 \\0 & y_{1} & 0 & y_{2} & 0\end{array}||} \\ & \mathrm{=\left|\frac{1}{2}\left[-2 \sqrt{2} y_{1}+2 \sqrt{2} y_{2}\right]\right|=\sqrt{2}\left|y_{2}-y_{1}\right|=\frac{(\sqrt{2}) \sqrt{12+16}}{111} }\\ & \mathrm{=\sqrt{56}=2 \sqrt{14}} \end{aligned}

 

Posted by

Shailly goel

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