Q. 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) $(x + y)^{2} = 2(x - y + 2)$
Explanation
(a)
As distance of vertex from origin is $\sqrt{2}\$and focus is$\ 2\sqrt{2}$
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$\therefore V(1,1)$ and $F(2,2)\ (ie,$ lying on $y = x)$
Length of latusrectum $= 4a = 4\sqrt{2}$ [where $a = \sqrt{2\rbrack}$
$\therefore$ By definition of parabola
$PM^{2} = (4a)(PN)$
where, $PN$ is length of perpendicular upon $x + y - 2 = 0$ ($ie$, tangent at vertex)
$\Rightarrow \ \frac{(x - y)^{2}}{2} = 4\sqrt{2}\left( \frac{x + y - 2}{\sqrt{2}} \right)$
$\Rightarrow \ (x - y)^{2} = 8(x + y - 2)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.