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Let $P\:\:and\:\:Q$ be points $(4,-4)\:\: and\:\: (9, 6)$ of the parabola $y^{2} = 4a( x - b)$ . Let $R$ be a point on the arc of the parabola between $P\:\:and\:\:Q$ . Then the area of $\Delta PRQ$ is largest when
Option: 1
$\angle PRQ=90^{0}$

Option: 2
$the \:\:point\:\:R\:is \:\:(4,4)$

Option: 3
$the\:\:point\:\:R\:\:is \:\:\left ( \frac{1}{4},1 \right )$

Option: 4
$None\:\:of\:\:these$

Answers (1)

best_answer

Parametric coordinates of parabola -

$x= at^{2}$

$y= 2at$

- wherein

For the parabola.

$y^{2}=4ax$

Since $(4,-4)\:\:and\:\:(9,6)$ lie on $y^{2}=4a(x-b)$

$\therefore 4=a(4-b)\:\:and\:\:9=a(9-b)$

$\therefore a=1\:\:and\:\:b=0$

$\therefore the\:\:parabola\:\:is\:\:y^{2}=4x$

let the point $R$ be $(t^{2},2t)$

$\therefore Area \:\:of\:\:\Delta PRQ$

$=\frac{1}{2}\begin{vmatrix} 4 &-4 &1 \\ 9 & 6 &1 \\ t^{2}&2t &1 \end{vmatrix}$

$=\frac{1}{2}\begin{vmatrix} 4 &-4 &1 \\ 5 & 10 &0 \\ t^{2}-4&2t+4 &0 \end{vmatrix}$ $=\frac{1}{2}\begin{vmatrix} 5 &10 \\ t^{2}-4 & 2t+4 \end{vmatrix}$

$=\frac{1}{2}(10t+20-10t^{2}+40)$

$=5t^{2}+5t+30=-5(t^{2}-t+\frac{1}{4})+30+\frac{5}{4}$

$=-5\left ( t-\frac{1}{2} \right )^{2}+\frac{125}{4}$

$\therefore Area\:\:in\:\:largest\:\:when\:\:t=\frac{1}{2}$

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himanshu.meshram

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