Question

Columns 1, 2 and 3 contain conics, equations of tangents to the conics and points of contact, respectively.
​​​​​​$\begin{array}{ccc}Column1& Column2& Column3\\ \left(I\right)x^2+y^2=a^2& \left(i\right)my=m^{2}x+a& \left(P\right)(\frac{a}{m^2},\frac{2a}{m})\\ \left(II\right)x^2+a^2{y}^2=a^2& \left(ii\right)y=mx+a\sqrt{m^2+1}& \left(Q\right)(\frac{-ma}{\sqrt{m^2+1}},\frac{a}{\sqrt{m^2+1}})\\ \left(III\right)y^2=4ax& \left(iii\right)y=mx+\sqrt{a^2{m}^2-1}& \left(R\right)(\frac{-a^{2}m}{\sqrt{a^2{m}^2+1}},\frac{1}{\sqrt{a^2{m}^2+1}})\\ \left(IV\right)x^2-a^2{y}^2=a^2& \left(iv\right)y=mx+\sqrt{a^2{m}^2+1}& \left(S\right)(\frac{-a^{2}m}{\sqrt{a^2{m}^2-1}},\frac{-1}{\sqrt{a^2{m}^2-1}})\end{array}$

If a tangent to a suitable conic (Column 1) is found to be $y=x+8$ and its point of contact is $(8,16)$, then which of the following options is the only CORRECT combination?

• A. $\left(I\right)\left(ii\right)\left(Q\right)$
• B. $\left(II\right)\left(iv\right)\left(R\right)$
• C. $\left(III\right)\left(i\right)\left(P\right)$
• D. $\left(III\right)\left(ii\right)\left(Q\right)$

Answer 1

The correct option is C $\left(III\right)\left(i\right)\left(P\right)$

Tangent at $(8,16)$ is $y=x+8$
Slope of tangent is positive hence possible curve will be
$y^2=4ax$
Therefore, equation of tangent is $my=m^{2}x+a$ and point of contact is $(\frac{a}{m^2},\frac{2a}{m})$

The correct combination of curve, tangent and point of contact is given in the table below:
$\begin{array}{ccc}\left(I\right)& \left(ii\right)& \left(Q\right)\\ \left(II\right)& \left(iv\right)& \left(R\right)\\ \left(III\right)& \left(i\right)& \left(P\right)\\ \left(IV\right)& \left(iii\right)& \left(S\right)\end{array}$