Question

$\text{List I}$ has four entries and $\text{List II}$ has five entries. Each entry of $\text{List I}$ is to be matched with one or more than one entries of $\text{List II}.$

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(A)}& \text{Tangents are drawn from the point}(2,3)& \text{(P)}& (9,-6)\\ & \text{to the parabola}{y}^2=4x.\text{Then point(s) of}& & \\ & \text{contact is (are)}& & \\ \text{(B)}& \text{From a point}P\text{on the circle}{x}^2+y^2=5,& \text{(Q)}& (1,2)\\ & \text{the equation of chord of contact to the}& & \\ & \text{parabola}{y}^2=4x\text{is}y=2(x-2).\text{Then}& & \\ & \text{the coordinates of}P\text{are}& & \\ \text{(C)}& P(4,-4),Q\text{are points on parabola}& \text{(R)}& (-2,1)\\ & y^2=4x\text{such that area of}△POQ\text{is}6& & \\ & \text{sq. units where}O\text{is the vertex. Then}& & \\ & \text{coordinates of}Q\text{may be}& & \\ \text{(D)}& \text{The common chord of circle}{x}^2+y^2=5& \text{(S)}& (4,4)\\ & \text{and parabola}6y=5x^2+7x\text{will pass}& & \\ & \text{through point(s)}& & \\ & & \text{(T)}& (-2,2)\end{array}$

Which of the following is the only CORRECT combination?

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

Answer 1

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

$\left(C\right)$
Let $Q(t^2,2t)$ be a point on the parabola.
Now, the area of triangle $OPQ$ is
$\left|\frac{1}{2}\left|\begin{array}{cc}0& 0\\ 4& -4\\ t^2& 2t\\ 0& 0\end{array}\right|\right|=6\Rightarrow 8t+4t^2=\pm 12\Rightarrow t^2+2t+1=1\pm 3\Rightarrow (t+1{)}^2=-2$ or $4$
$\Rightarrow (t+1{)}^2=4\Rightarrow t=-1\pm 2\Rightarrow t=-3,1$
Then, the point $Q$ is $(1,2)$ or $(9,-6)$.
$\left(C\right)\to \left(P\right),\left(Q\right)$

$\left(D\right)$
Points $(1,2)$ and $(-2,1)$ satisfy both the curves.
$\left(D\right)\to \left(Q\right),\left(R\right)$