Question

Show that the line $x+2y+8=0$ is tangent to the parabola $y^2=8x$. Hence find the point of contact

Answer 1

Given line is
$x+2y+8=0$ ..... (i)
and equation of parabola
$y^2=8x$ ..... (ii)
Now for (i),
$x=-8-2y$
Putting the value of $x$ in equation (ii),
$y^2=8(-8-2y)$
$\Rightarrow y^2=-64-16y$
$\Rightarrow y^2+16y+64=0$
$\Rightarrow y^2+8y+8y+64=0$
$\Rightarrow y(y+8)+8(y+8)=0$
$\Rightarrow (y+8)(y+8)=0$
$\Rightarrow (y+8{)}^2=0$
The quadratic equation in $y$ has two equal roots each being $-8$
Hence the given line touches the parabola.
When $y=-8$
From (i),
$x=-8-2(-8)$
$=-8+16=8$
$\therefore$ the point of contact is $(8,-8)$.