Question

The vertex of a parabola is $(a,0)$ and the directrix is $x+y=3a$. The equation of the parabola is

• A. $x^2-2xy+y^2+6ax+10ay-7a^2=0$
• B. $x^2+2xy+y^2+6ax+10ay+2a^2=0$
• C. $x^2+2xy+y^2+6ax+10ay=2a^2$
• D. None of these

Answer 1

The correct option is A $x^2-2xy+y^2+6ax+10ay-7a^2=0$

Equation of axis is given by, $x-y=c$, since the directrix and axis of the parabola are perpendicular to each other.

It also passes through vertex $(a,0)\Rightarrow c=a$

Now solving directrix and axis to get foot of directrix. $x=2a,y=a$

We know vertex is mid point of foot of directrix and focus. $\therefore$

focus is $S(0,-a)$

Now using definition of parabola,
$P{S}^2=P{M}^2\Rightarrow (x-0{)}^2+(y+a{)}^2={\left(\frac{x+y-3a}{\sqrt{2}}\right)}^2$

$\Rightarrow 2(x^2+y^2+2ay+a^2)=x^2+y^2+9a^2+2xy-6ax-6ay$

$\Rightarrow x^2+y^2-2xy+6ax+10ay-7a^2=0$