Browse Questions

Show that normal to the parabola $y^2=4x$ at the point (1,2) meet it again at some point of parabola, find the point?

$\begin{array}{1 1}(a)\;(9,-6)\\(b)\;(8,-6)\\(c)\;(7,-6)\\(d)\;(9,6)\end{array}$

We know normal to the parabola $y^2=4x$ is
$y-y_1=-\large\frac{-y_1}{2a}$$(x-x_1) (y-2)=-\large\frac{2}{2\times 1}$$(x-1)$
$y+x=3$
$y^2=4x$
$y^2=4(3-y)$
$y^2+4y-12=0$
$y^2-2y+6y-12=0$
$y(y-2)+6(y-2)=0$
$(y-2)(y+6)=0$
$y=2,-6$
$3-y=x$
$3+6=x$
$x=9$
Hence (9,-6) is another point on the parabola.
Hence (a) is the correct answer.