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The tangents at the points $\mathrm{A(1,3)\: and\: B(1,-1)}$ on the parabola $\mathrm{y^{2}-2 x-2 y=1}$ meet at the point $\mathrm{P}$ . Then the area $\mathrm{(in\: unit { }^{2} )}$ of the triangle $\mathrm{P A B}$ is:
Option: 1
$4$

Option: 2
$6$

Option: 3
$7$

Option: 4
$8$

Answers (1)

best_answer

$\mathrm{y^{2}-2 x-2 y=1 } \\$

$\mathrm{\Rightarrow y^{2}-2 y=2 x+1} \\$

$\mathrm{\Rightarrow (y-1)^{2}=2(x+1)}$

$\mathrm{Let\; P\; be \;(h, k)}\\$

$\mathrm{Chord\; of\; Contact: T=0\quad }\\$

$\mathrm{k y-2\left(\frac{x+h}{2}\right)-2\left(\frac{y+k}{2}\right)=1}$

$\mathrm{\Rightarrow \quad k y-x-h-y-k=1}$

$\mathrm{\Rightarrow \quad-x+y(k-1)-h-k-1=0}$

This should be same as $\mathrm{x=1}$

$\mathrm{\therefore k-1=0 \Rightarrow k=1} \\$

$\mathrm{\text { and }-h-k-1=1} \\$

$\mathrm{\Rightarrow -h-1-1=1} \\$

$\mathrm{\Rightarrow h=-3} \\$

$\mathrm{\therefore \quad P(-3,1)} \\$

$\mathrm{\text { Area of } P A B=\mid\frac{1}{2}\left|\begin{array}{ccc} -3 & 1 & 1 \\ 1 & 3 & 1 \\ 1 & -1 & 1 \end{array}\right| \mid} \\$

$=8$

Hence correct option is 4

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