Question

Match $\text{List I}$ with the $\text{List II}$ and select the correct answer using the code given below the lists :

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(A)}& \text{Radius of the largest circle which passes through the focus of the parabola}{y}^2=4x\text{and is completely}& \text{(P)}& 16\\ & \text{contained in it, is}& & \\ \text{(B)}& \text{If the shortest distance between the curves}{y}^2=4x\text{and}{y}^2=2x–6\text{is}d\text{, then}{d}^2\text{is}& \text{(Q)}& 5\\ \text{(C)}& \text{Let}AB\text{be a focal chord of}{y}^2=12x\text{with focus}S.\text{The harmonic mean of lengths of segments}AS& \text{(R)}& 6\\ & \text{and}BS\text{is}& & \\ \text{(D)}& \text{Tangents drawn from}P\text{meet the parabola}{y}^2=16x\text{at}A\text{and}B.\text{If these two tangents are}& \text{(S)}& 4\\ & \text{perpendicular, then the least value of}\sqrt{AB}\text{is}& & \end{array}$


Which of the following is CORRECT ?

• A. $\left(A\right)\to \left(R\right),\left(B\right)\to \left(S\right)$
• B. $\left(A\right)\to \left(S\right),\left(B\right)\to \left(Q\right)$
• C. $\left(A\right)\to \left(P\right),\left(B\right)\to \left(Q\right)$
• D. $\left(A\right)\to \left(S\right),\left(B\right)\to \left(R\right)$

Answer 1

The correct option is B $\left(A\right)\to \left(S\right),\left(B\right)\to \left(Q\right)$

$\left(A\right)$
Let $(h,0)$ be the center of circle.

Here, $h=1+r$
Equation of the circle is,$(x-h{)}^2+y^2=r^2$
$\Rightarrow (x-r-1{)}^2+y^2=r^2$
$\Rightarrow x^2+y^2-2x+2r-2xr+1=0$
Above circle touches the curve $y^2=4x,$
$x^2+2x-2xr+2r+1=0$
$x^2+x(2-2r)+2r+1=0$
The above quadratic equation has equal roots, because the circle touches the parabola from inside and both the curves are symmetrical w.r.t. $x-$axis
$\Rightarrow (2-2r{)}^2=4(2r+1)$
$\Rightarrow 4+4r^2-8r=8r+4$
$\Rightarrow 4r^2-16r=0$
$\Rightarrow r^2-4r=0$
$\Rightarrow r(r-4)=0$
Radius cannot be zero. So, $r=4$

$\left(B\right)$
Common normal to
$y^2=4x$ and $y^2=2x-6$
Equation of normal to the $y^2=2(x-3)$ is :
$y=m(x-3)-m-\frac{1}{2}{m}^3\dots \left(1\right)$

Equation of normal to the $y^2=4x$ is
$y=mx-2m-m^3\dots \left(2\right)$
Here, equation $\left(1\right)$ and $\left(2\right)$ are indentical
So, $-4m-\frac{1}{2}{m}^3=-2m-m^3$
$\frac{m^3}{2}-2m=0\Rightarrow m=0,m=\pm 2$
Then points on curves are $P(m^2,-2m)$ and $Q(\frac{m^2}{2}+3,-m)$
For $m=0$
$P(0,0),Q(3,0)$ and $d=3$
For $m=2$
$P(4,-4),Q(5,-2)$ and $d=\sqrt{5}$
For $m=-2$
$P(4,4),Q(5,2)$ and $d=\sqrt{5}$
Hence, shortest distance $d=\sqrt{5}\Rightarrow d^2=5$