Question

Axis of a parabola is $y=x$ and vertex and focus are at distances $\sqrt{2}$ and $2\sqrt{2}$, respectively, from the origin. Then equation of the parabola is

• A. $(x-y{)}^2=8(x+y-2)$
• B. $(x+y{)}^2=2(x+y-2)$
• C. $(x-y{)}^2=4(x+y-2)$
• D. $(x+y{)}^2=2(x-y+2)$

Answer 1

Answer: (A)

From the figure shown here, we have $OV=\sqrt{2}$ and $OS=2\sqrt{2}$. The vertex $V$ is obtained as follows:
$\left(V(OV)\cos 45^{\circ },(OV)\sin 45^{\circ }\right)=\left(\sqrt{2}\frac{1}{\sqrt{2}},\sqrt{2}\frac{1}{\sqrt{2}}\right)=(1,1)$
$S=\left((OS)\cos 45^{\circ },(OS)\sin 45^{\circ }\right)$
$S=(2,2)$

The equation of directrix is $OV=VS$
where
$OV=\sqrt{2}$
$VS=\sqrt{(2-1{)}^2+(2-1{)}^2}=\sqrt{2}$
Now, $a=\sqrt{2}$ and therefore,
$x+y=0$ (directrix)
Equation of parabola: we have $P(x,y)$ and
$PS=PM$
$\sqrt{(x-2{)}^2+(y-2{)}^2}=\frac{|x+y|}{\sqrt{2}}$
$(x+y{)}^2=2\left((x-2{)}^2+(y-2{)}^2\right)$
$x^2+y^2+2xy=2\left(x^2+y^2\right)+16-8x-8y$
$x^2+y^2-2xy=8x+8y-16$
$(x-y{)}^2=8(x+y-2)$