Question

A parabola has its vertex and focus in the first quadrant and axis along the line $y=x$. If the distances of the vertex and focus from the origin are respectively $\sqrt{2}$ and $2\sqrt{2}$, then an equation of the parabola is

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

Answer 1

The correct option is C ${(x-y)}^2=8(x+y-2)$

We have,

The line $y=x......\left(1\right)$

Then, $y=mx$

On compare that,

$m=1$

$m=\tan \theta =\tan {45}^o$

$\theta ={45}^o$

Now,

Let the vertex of parabola is

$\begin{align}

$(x\cos \alpha ,x\sin \alpha )$

$=(\sqrt{2}\cos {45}^o,\sqrt{2}\sin {45}^o)$

$=(1,1)$

Let the focus of parabola is

$(y\cos \alpha ,y\sin \alpha )$

$\Rightarrow (y\cos {45}^o,y\sin {45}^o)$

$\Rightarrow (2\sqrt{2}\cos {45}^o,2\sqrt{2}\sin {45}^o)$

$\Rightarrow (2,2)$

Now, equation of parabola

$\sqrt{{(x-2)}^2+{(y-2)}^2}=\left|\frac{y+x}{\sqrt{2}}\right|$

On squaring both side and we get,

$x^2+4-4x+y^2+4-4y=\frac{x^2+y^2+2xy}{2}$

$\Rightarrow 2x^2+2y^2-8x-8y+16=x^2+y^2+2xy$

$\Rightarrow x^2+y^2-2xy=8x+8y-16$

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

Hence, this is the required answer.