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$\\On\:parabola\:y^2=4x\:three\:point\:A\left(4,-4\right),\:B\left(9,\:6\right)\:\\and\:C\:lies\:on\:it\:then\:area\:of\:triangle\:ABC\left(max.\right)\:is-$

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$y^2 = 4x$

Any point on the parabola is

$(at^2, 2at)$

Here, $a = 1$

So, the three points of triangles are : (4, -4), (9, 6) and $(at^2, 2at)$ .

So the Area :

$\Delta=\frac{1}{2}\left|\begin{array}{lll}{x_{1}} & {y_{1}} & {1} \\ {x_{2}} & {y_{2}} & {1} \\ {x_{3}} & {y_{3}} & {1}\end{array}\right|$

$\Delta=\frac{1}{2}\left|\begin{array}{lll}{4} & {-4} & {1} \\ {9} & {6} & {1} \\ {at^2} & {2at} & {1}\end{array}\right|$

$\Delta=\frac{1}{2}[4(6-2at)+4(9-at^2)+1(18at-6at^2)]$

$\Delta=\frac{1}{2}[24-8at+36-4at^2+18at-6at^2]$

$\Delta=\frac{1}{2}[-10at^2+10at+60]$

$\Delta=5[-at^2+at+6]$

$\Delta=5\left [ \left ( t-\frac{1}{2} \right )^2-\frac{25}{6} \right ]$

As we can see Area will be maximum when $t=\frac{1}{2}$

So,

Maximum Area = $5\times\frac{25}{6}=\frac{125}{6}$ .

Posted by

Pankaj Sanodiya

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