Question

Axis of a parabola is $y=x$ and vertex and focus are at a distance $\sqrt{2}$ 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

The correct option is A

${(x–y)}^2=8(x+y–2)$

Explanation for correct option:

Step-1: Draw the figure with the help of given data.

Since, given vertex is at $\sqrt{2}$ and focus is at $2\sqrt{2}$ from the origin therefore we can consider coordinate of the vertex is $\left(1,1\right)$

and coordinate of the focus is $\left(2,2\right)$

equation of directrix is $x+y=0$

Step-2: Finding equation of parabola.

Let, variable point on parabola is $\left(h,k\right)$.

According to definition.

Distance from variable point to focus $=$Distance from variable point to directrix.

$\Rightarrow$ $\sqrt{{\left(h-2\right)}^2+{\left(k-2\right)}^2}=\frac{\left|h+k\right|}{\sqrt{1^2+1^2}}$

$\Rightarrow$ $h^2+k^2-4h-4k+8=\frac{h^2+k^2+2hk}{2}$

$\Rightarrow$ $h^2+k^2-2hk=8h+8k-16$

$\Rightarrow$ ${\left(h-k\right)}^2=8\left(h+k-2\right)$

Step-3: Replace $\left(h,k\right)\mathrm{to}\left(x,y\right)$

$\Rightarrow$${\left(x-y\right)}^2=8\left(x+y-2\right)$

Hence, correct answer is option $\left(A\right)$.