Question

Tangents to the parabola at the extremities of a common chord $AB$ of the circle $x^2+y^2=5$ and the parabola $y^2=4x$ intersect at the point $T$. A square $ABCD$ is constructed on this chord lying inside the parabola, then $\frac{{[{\left(TC\right)}^2+{\left(TD\right)}^2]}^2}{1600}$ is equal to

Answer 1

Given circle $x^2+y^2=5$ and the parabola $y^2=4x$ have the common chord $AB$.
Solving above two equations,
$x^2+4x-5=0$$\Rightarrow x=1or-5$
but $x\ne -5$ as parabola lies in quad I and IV
at $x=1$ $y=\pm 1$
$\therefore$ $A$ is $(1,2)$ and $B$ is $(1,-2)$
Equations of tangents to parabola at $A$ and $B$ are $x-y+1=0$ and $x+y+1=0$, respectively
Solving for point of intersection gives $T(-1,0)$.
As square $ABCD$ is drawn on $AB$, which lies inside parabola, co-ordinates of $C$ and $D$ are $(5,-2)$ and $(5,2)$ respectively.
$\frac{{[{\left(TC\right)}^2+{\left(TD\right)}^2]}^2}{1600}$$=\frac{(40+40{)}^2}{1600}=\frac{6400}{1600}=4$