Question

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

$\begin{array}{ccccc}& \text{List I}& & \text{List II}& \\ \text{(A)}& \text{A possible point of intersection of the}& & & \\ & \text{parabola}{y}^2=4x\text{and the circle having}& & & \\ & \text{centre at}(6,5),\text{which cut orthogonally, is}& \text{(P)}& (6,3)& \\ \text{(B)}& \text{From a point}P,\text{tangents}PQ\text{and}PR& & & \\ & \text{are drawn to the ellipse}{x}^2+2y^2=2,& & & \\ & \text{so that the equation of}QR\text{is}x+3y=1.& & & \\ & \text{Then the coordinates of}P\text{are}& \text{(Q)}& (4,4)& \\ \text{(C)}& \text{The vertices of the hyperbola}& & & \\ & 9x^2-16y^2-36x+96y-252=0\text{are}& \text{(R)}& (-2,3)& \\ \text{(D)}& \text{A point on}{y}^2=4x\text{at which the normal}& & & \\ & \text{makes equal angles with the coordinate}& & & \\ & \text{axes is}& \text{(S)}& (1,-2)& \\ & & \text{(T)}& (2,3)& \end{array}$

Which of the following is CORRECT combination?

• A. $A\to Q;B\to T$
• B. $A\to R,S;B\to P$
• C. $A\to R;B\to P,T$
• D. $A\to S;B\to P$

Answer 1

The correct option is A $A\to Q;B\to T$

$\left(A\right)$
Given parabola is $y^2=4x$
Any point on it can be written as $(t^2,2t)$
Equation of tangent at $(t^2,2t)$ to the parabola is
$y\left(2t\right)=2(x+t^2)\Rightarrow yt=x+t^2$

As the parabola and circle cut orthogonally, the tangent on the parabola must pass through the centre of the circle $(6,5)$, so
$5t=6+t^2\Rightarrow t^2-5t+6=0\Rightarrow t=2,3$
Hence, possible points of intersection are $(4,4)$ and $(9,6).$
$A\to Q$

$\left(B\right)$
Given ellipse is $\frac{x^2}{2}+y^2=1$
Let the point $P=(h,k)$,

Equation of the chord of contact $QR$ drawn from the point $P$ to the ellipse is
$T=0\Rightarrow hx+2ky-2=0$
Given equation of $QR$ is $x+3y-1=0$
Comparing both equations, we get
$\frac{h}{1}=\frac{2k}{3}=\frac{-2}{-1}\Rightarrow h=2,k=3\therefore P\equiv (2,3)$
$B\to T$