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 $O V=\sqrt{2}$ and $O S=2 \sqrt{2}$. 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}+2 x y=2\left(x^{2}+y^{2}\right)+16-8 x-8 y $
$x^{2}+y^{2}-2 x y=8 x+8 y-16$
$(x-y)^{2}=8(x+y-2)$