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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
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Verified Answer
The correct answer is:
$(x-y)^2=8(x+y-2)$
$(x-y)^2=8(x+y-2)$
As distance of vertex from origin is $\sqrt{2}$ and focus is $2 \sqrt{2}$. $\therefore V(1,1)$ and $F(2,2)$
[i.e. lying on $y=x$ ] where, length of latusrectum
$$
\begin{aligned}
& =4 a, \\
& =4 \sqrt{2}
\end{aligned}
$$
$$
[\because \text { where } a=\sqrt{2}]
$$
$\therefore$ By definition of parabola
$$
P M^2=(4 a)(P N)
$$
where, $P N$ is length of perpendicular upon $x+y-2=0$
[i.e. tangent at vertex]
$$
\begin{array}{ll}
\Rightarrow & \frac{(x-y)^2}{2}=4 \sqrt{2}\left(\frac{x+y-2}{\sqrt{2}}\right) \\
\therefore & (x-y)^2=8(x+y-2)
\end{array}
$$
Hence, the equation of parabola is $(x-y)^2=8(x+y-2)$.
[i.e. lying on $y=x$ ] where, length of latusrectum
$$
\begin{aligned}
& =4 a, \\
& =4 \sqrt{2}
\end{aligned}
$$
$$
[\because \text { where } a=\sqrt{2}]
$$
$\therefore$ By definition of parabola
$$
P M^2=(4 a)(P N)
$$
where, $P N$ is length of perpendicular upon $x+y-2=0$
[i.e. tangent at vertex]
$$
\begin{array}{ll}
\Rightarrow & \frac{(x-y)^2}{2}=4 \sqrt{2}\left(\frac{x+y-2}{\sqrt{2}}\right) \\
\therefore & (x-y)^2=8(x+y-2)
\end{array}
$$
Hence, the equation of parabola is $(x-y)^2=8(x+y-2)$.
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