Question

$A$ is the vertex and $S$ is focus of the parabola $y^2=4ax.$ A circle with centre at $A$ and radius $\frac{3}{2}a$ cuts the parabola in $P$ and $Q.$ Find the equation of $PQ.$

• A. $x=\frac{a}{2}$
• B. $x=\frac{a}{4}$
• C. $x=\frac{a}{3}$
• D. $x=\frac{a}{9}$

Answer 1

The correct option is A $x=\frac{a}{2}$

Parabola is $y^2-4ax=0$
Circle is $x^2+y^2-{\left(\frac{3}{2}a\right)}^2=0$
If they intersect at $P$ and $Q,$ then solving them we get
or $x^2+4ax-\frac{9}{4}{a}^2=0$
or $4x^2+16ax-9a^2=0$
$(2x+9a)(2x-a)=0\therefore x=\frac{a}{2}$
Note : $x$ cannot be negative because otherwise $y$ will be imaginary from parabola.
Putting $x=\frac{a}{2},$ we get $y^2=2a^2\therefore y=\pm a\sqrt{2}$
$\therefore$ $P$ is $(\frac{a}{2},a\sqrt{2})$ and $Q$ is $(\frac{a}{2},-a\sqrt{2})$
Thus equation of $PQ$ is, $x=\frac{a}{2}$