Question

If lines $x-y+2=0$ and $2x-y-2=0$ meet at a point $P,$ then equation of tangent drawn to the parabola $y^2=8x$ from the point $P$ is

• A. $x-2y+8=0$
• B. $x+y-16=0$
• C. $3x-y-16=0$
• D. $x-3y+16=0$

Answer 1

The correct option is A $x-2y+8=0$

$\begin{array}{c}x-y+2=0\\ 2x-y-2=0\end{array}\}$
Solving the two equations, we get
$x=4$ and $y=6$
So, the tangent passes through $(4,6)$
Let slope of the tangent be $m.$
$y-6=m(x-4)$
$\Rightarrow y=mx+6-4m$ which is tangent to the parabola $y^2=8x$
$\therefore 6-4m=\frac{2}{m}$ $[\text{Using}c=\frac{a}{m}]$
$\Rightarrow 3m-2m^2=1$
$\Rightarrow 2m^2-3m+1=0$
$\Rightarrow m=\frac{1}{2},1$

For $m=\frac{1}{2},$
$y=\frac{1}{2}x+6-2$
$\Rightarrow x-2y+8=0$

For $m=1,$
$y=x+2$ that is the given line itself.
So, tangent is $:x-2y+8=0$