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

Answer 1

The correct option is A $\left(C\right)\to \left(R\right),\left(D\right)\to \left(S\right)$

$\left(R\right)$
Equation of parabola is $y^2=12x$
Length of latus rectum
$=4a=12\Rightarrow a=\frac{12}{4}=3$

From the figure, we can say that $S(3,0)$
is the focus and $AB$ is the focal chord of the parabola.
Let $A(3t_1^2,6t_1)$ and
$B(3a{t}_2^2,6a{t}_2)$.
Hence from the figure, we get
$AS=3t_1^2+3,BS=3t_2^2+3$
$\therefore$Harmonic mean
$\frac{2}{\frac{1}{AS}+\frac{1}{BS}}=\frac{2}{\frac{1}{3(t_1^2+1)}+\frac{1}{3(t_2^2+1)}}=\frac{2\times 3}{\frac{1}{(t_1^2+1)}+\frac{1}{(t_2^2+1)}}$
$\therefore$We get the harmonic mean as
$\frac{6}{\frac{1}{(t_1^2+1)}+\frac{1}{(t_2^2+1)}}$
Here, slope of $AS=$ slope of $BS$
Thus, for a focal chord $t_1{t}_2=-1$ thus,
$t_2=-\frac{1}{t_1}$
Now, harmonic mean
$=\frac{6}{\frac{1}{t_1^2+1}+\frac{1}{t_2^2+1}}=\frac{6}{\frac{1}{t_1^2+1}+\frac{t^2}{1+t_1^2}}=\frac{6(1+t^2)}{(1+t^2)}=6$


$\left(S\right)$
Equation of parabola is
$y^2=16x$.
For $y^2=4ax\Rightarrow a=4$
As given that tangents are perpendicular. So, $AB$ is the focal chord of the parabola.
Let $A(a{t}_1^2,2a{t}_1),B(a{t}_2^2,2a{t}_2)$.
Then length of focal chord is $AB=a{(t+\frac{1}{t})}^2$
Minimum value of $AB$ is $a\times 2^2=4\times 4$
$\Rightarrow \sqrt{AB}=4$