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

Since, distance of vertex from origin is $\sqrt{2}$ and focus is $2\sqrt{2}$, there are four possible parabolas that can be constructed. Parabola with vertex and focus both in the first quadrant, vertex and focus both in the third quadrant, vertex in the first quadrant while focus in the third quadrant and vice versa.
Considering the case where both the focus and the vertex are in the first quadrant we get $V(1,1)$ and $F(2,2)$ (i.e. lying on y=x)
Let $P(h,k)$ be any point on the parabola.
Equation of directrix of the parabola is $x+y=0$
Using the definition of parabola, distance of $P(h,k)$ from directrix is equal to its distance from focus.
$\sqrt{(h-2{)}^2+(k-2{)}^2}=|\frac{h+k}{\sqrt{2}}|\Rightarrow h^2+4-4h+k^2+4-4k=\frac{(h+k{)}^2}{2}\Rightarrow 2h^2+2k^2+16-8(h+k)=(h+k{)}^2\Rightarrow h^2+k^2-2hk=8(h+k)-16\Rightarrow (h-k{)}^2=8(h+k-2)\therefore \text{equation of the parabola is}(x-y{)}^2=8(x+y-2)$