Question

Find the equation of the parabola whose vertex and focus lie on the $x$ axis at distances $a$ and $a_1$ from the origin respectively.

• A. $y^2=4(-a_1+a)(x-a).$
• B. $y^2=4(a_1-a)(x+a).$
• C. $y^2=4(a_1-a)(x-a).$
• D. $y^2=4(a_1+a)(x+a).$

Answer 1

The correct option is C $y^2=4(a_1-a)(x-a).$

$A$ is $(a,0),S=(a_1,0)$
$\therefore AS=a_1-a=A$
$\therefore L.R.=4AS=4(a_1-a).$
Since the axis is the $x-$axis and vertex is $(a,0),$ hence by definition its equation is
$Y^2=4AX$

where $4A=4AS=4(a_1-a)$.
or ${(y-0)}^2=4(a_1-a)(x-a)$
or $y^2=4(a_1-a)(x-a).$
Ans: C