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# The line $2x+3y+9=0$ touches the parabola $y^{2}=8x$ at the point

$\begin{array}{1 1}(1)(0 , -3)&(2)(2 , 4)\\(3)(-6 , \frac{9}{2})&(4)(\frac{9}{2} ,-6 )\end{array}$

The equation of the parabola is $y^2=8x$
$y^2=4(2) x => a =2$
The equation of the given line is
$2x+3y+9=0$
$3y=-2x-9$
$y= -\large\frac{2}{3} x -\frac{9}{3}$
$y= -\large\frac{2}{3} x$$-3 m=-\large\frac{2}{3}$$x -3$
$c=-3$
The point of contact is $\bigg( \large\frac{2}{(\Large\frac{-2}{3} )^2}. \frac{2 \times 2}{(\Large\frac{-2}{3})} \bigg)$
$\qquad=\bigg( \large\frac{2}{\Large\frac{4}{9} }. \frac{4}{-2}$$\times 3 \bigg)$
$\qquad= \bigg (\large\frac{9}{2}, -6 \bigg)$
Hence 4 is the correct answer.