An equation of a tangent to the parabola $y^2=8x$ is $y=x+2$. the point on this line from which the other tangent Problem : 

An equation of a tangent to the parabola $y^2=8x$ is $y=x+2$. the point on this line from which the other tangent to the parabola is perpendicular to the  given tangent is given by ...

Solution : 

Slope at any point on the parabola from where tangent can be drawn can be taken by differentiating equation of parabola $y^2=8x$ 

Which gives $2y \frac{dy}{dx}=8 \Rightarrow \frac{dy}{dx}=4/y$

Slope of the given line is equal to the slope of the line at this point therefore, $y =4$ and $x=2$.

Please suggest whether this is the correct method of doing this and how can I proceed this problem further thanks..
Hints:


You know the gradient of the tangent, so you can find the slope of a line perpendicular to it.
You can use the same method you have already used to find the point where this second line is tangent to the parabola and so the equation of the second tangent.
You now have the equation for two straight lines, and find where they intersect.
$1)$ $y^2=8x$

$\frac{dy}{dx} = 8x$

 8---------(1


$\frac{dy}{dx}= y(2)$

$2(2)= 4$

$y-2=4(x-8)$
$y-2= 4x-32$

$y= 4x-16$    Tangent of the line.