Question

Write down the equations to the tangent and normal at the end points of the latus rectum of the parabola $y^2=12x$.

Answer 1

$y^2=12x4a=12\Rightarrow a=3$

The end points of latus rectum are $(a,2a)$ and $(a,-2a)$

For given parabola end points are $L(3,6)$ and $L^{\prime }(3,-6)$

Equation of tangent at $L$ is

$y\left(6\right)=12\left(\frac{x+3}{2}\right)y=x+3$

Slope of tangent $=m=1$

$\Rightarrow$ Slope of normal $=-1$

Equation of normal is

$y-6=-1(x-3)y-6=-x+3x+y=9$

Equation of tangent at $L^{\prime }$ is

$y(-6)=12\left(\frac{x+3}{2}\right)-y=x+3x+y+3=0$

Slope of tangent $=m=-1$

$\Rightarrow$ Slope of normal $=1$

Equation of normal is

$y-(-6)=1(x-3)y+6=x-3x-y=9$