Question

Write down the equations to the tangent and normal at the ends of the latus rectum of the parabola $y^2=4a(x-a).$

Answer 1

$y^2=4a(x-a)$

Focus of the parabola is $(2a,0)$

End points of latus rectum are $L(2a,2a)$ and $L^{\prime }(2a,-2a)$

$y^2=4a(x-a)y^2=4ax-4a^2$

Equation of tangent at $L$ is

$y\left(2a\right)=4a\left(\frac{x+2a}{2}\right)-4a^{2}2ay=2a(x+2a)-4a^{2}y=x+2a-2ay=x$

Slope of tangent $=m=1$

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

Equation of normal is

$y-2a=-1(x-2a)y-2a=-x+2ax+y=4a$

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

$y(-2a)=4a\left(\frac{x+2a}{2}\right)-4a^2-2ay=2a(x+2a)-4a^2-y=x+2a-2ay=-xy+x=0$

Slope of tangent is $m=-1$

$\Rightarrow$ Slope of normal $=1$

Then, equation of normal is

$y-(-2a)=1(x-2a)y+2a=x-2ax-y=4a$