Question

The length of latus rectum of the parabola x =ay2 + ay-a is

Answer 1

Given equation of parabola,

$\longrightarrow x=ay^2+ay-a$

We need to find length of latus rectum of this parabola.

So,

$\longrightarrow x=ay^2+ay-a$

$\longrightarrow x=a(y^2+y-1)$

$\longrightarrow x=a\left(y^2+y+\dfrac{1}{4}-\dfrac{5}{4}\right)$

$\longrightarrow x=a\left(y+\dfrac{1}{2}\right)^2-\dfrac{5a}{4}$

$\longrightarrow\left(y+\dfrac{1}{2}\right)^2=\dfrac{1}{a}\left(x+\dfrac{5a}{4}\right)$

Now our equation is in the form $(y')^2=4a'x'$ where,

• $y'=y+\dfrac{1}{2}$

• $4a'=\dfrac{1}{a}$

• $x'=x+\dfrac{5a}{4}$

We know the coefficient of $x$ in equation of parabola is the length of latus rectum.

Hence here, $\dfrac{1}{a}$ is the length of latus rectum.