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Q.
If the image of the parabola $y=x^{2}$ in the line $x+y=1$ is $x+y^{2}=k y$, then $k$ is equal to
Conic Sections
Solution:
General point on $y=x^{2}$ is $\left(t, t^{2}\right) .$
Let us find the image of $\left(t, t^{2}\right)$ in $4 y=2 x+3$
$\frac{a+t}{2}+\frac{b+t^{2}}{2}=1$
$a+b+b+t^{2}=2 $
and $\frac{b-t^{2}}{a-t}=1$
Solving we get $a+b^{2}=2 b$
$\Rightarrow x+y^{2}=2 y $
$ \Rightarrow k=2$
