Question

Consider a circle $S:{(x+2)}^2+{(y-8)}^2=16$ and a parabola $P:y^2=8x.TA$ and $TB$ are two tangents drawn from a point $T$ on the parabola $P=0$ to the circle $S=0$ such that $TA+TB$ is minimum. $A$ and $B$ are the points of contact.

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(A)}& \text{If}T\equiv (a,b),\text{then}(a+b)\text{equals}& \text{(P)}& 6\\ \text{(B)}& \text{Sum of ordinates of}A\text{and}B\text{is}& \text{(Q)}& 8\\ \text{(C)}& \text{Minimum value of}(TA+TB)\text{is}& \text{(R)}& 10\\ \text{(D)}& \text{Area of}\Delta TAB\text{is}& \text{(S)}& 12\end{array}$

Which of the following is a CORRECT combination ?

• A. $\text{(C)}\to \text{(S)},\text{(D)}\to \text{(R)}$
• B. $\text{(C)}\to \text{(R)},\text{(D)}\to \text{(S)}$
• C. $\text{(C)}\to \text{(P)},\text{(D)}\to \text{(Q)}$
• D. $\text{(C)}\to \text{(Q)},\text{(D)}\to \text{(Q)}$

Answer 1

The correct option is D $\text{(C)}\to \text{(Q)},\text{(D)}\to \text{(Q)}$


For $TA+TB$ to be minimum, $T$ should be along the common normal.
$\Rightarrow \frac{1}{t}\times \frac{4t-8}{2t^2+2}=-1$
$\Rightarrow 2t-4=-t(t^2+1)$
$\Rightarrow t^3+3t-4=0$
$\Rightarrow t=1$
$T\equiv (2,4)\Rightarrow a+b=6$
$\text{(A)}\to \text{(P)}$

$T$ lies on the director circle ${(x+2)}^2+{(y-8)}^2=32$
$\therefore TA\perp TB$
$\Rightarrow$ figure $CATB$ is a square.
Now tangent to circle $S=0$ is
$y-8=m(x+2)\pm 4\sqrt{1+m^2}$
Putting $(2,4)$ in the tangent,
$-4=4m\pm 4\sqrt{1+m^2}$
$\Rightarrow {(m+1)}^2=1+m^2$
$\Rightarrow m=\infty$ or $m=0$
$\Rightarrow$ Tangent is $x=2$ and $y=4$
$\therefore A\equiv (2,8)$ and $B\equiv (-2,4)$
$\text{(B)}\to \text{(S)}$

$(TA+TB{)}_{min}=8$
$\text{(C)}\to \text{(Q)}$

Area of $\Delta TAB$ $=\frac{1}{2}\times 4\times 4=8$
$\text{(D)}\to \text{(Q)}$