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Q.
If graph of the expression $y=x^{2}-8 x+12$ is shown in the figure then area of the square $A B C D$ inscribed between parabola and $x$-axis is given by
Conic Sections
Solution:
Let the side length be $2k$
Now point $((k+4),-2 k)$ lies on parabola $y=x^{2}-8 x+12$.
$\therefore (-2 k)=(k+4)^{2}-8(k+4)+12$
$\Rightarrow k^{2}+2 k-4=0$
$\Rightarrow k=\frac{-2+\sqrt{4+16}}{2}=-1+\sqrt{5}$
So, area $=4 k^{2}=4(-1+\sqrt{5})^{2}=4(6-2 \sqrt{5})$
