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  • Normal PO, PA and PB ('O' being the origin) are drawn to <img alt="\mathrm{y^{2}=4 x\: from \: P(h, 0)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7By%5E%7B2%7D%3D4%20x

Normal PO, PA and PB ('O' being the origin) are drawn to \mathrm{y^{2}=4 x\: from \: P(h, 0)}. If \mathrm{\angle \mathrm{AOB}=\pi / 2}, then area of quadrilateral \mathrm{\mathrm{OAPB}} is equal to

Option: 1

12 sq. units


Option: 2

24 sq. units


Option: 3

6 sq. units


Option: 4

18 sq. units


Answers (1)

best_answer

Let \mathrm{A \equiv\left(t^{2}, 2 t\right), B \equiv\left(t^{2},-2 t\right)}
\mathrm{m_{O A}=\frac{2}{t}, \quad m_{O B}=-\frac{2}{t}}
\mathrm{\Rightarrow \quad t^{2}=4}
\mathrm{\Rightarrow \quad t=2}


Equation of normal AP is
\mathrm{y=-2 x+4+8}
\mathrm{P \equiv(6,0)}

Thus area of quadrilateral
\mathrm{\text { OAPB } =\frac{1}{2}(O P)(A B) }
                \mathrm{=\frac{1}{2} .6 .8=24 \text { sq. units } }.

Posted by

shivangi.bhatnagar

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