Question

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

The tangent to a suitable conic (Column 1) at $(\sqrt{3},\frac{1}{2})$ is found to be $\sqrt{3}x+2y=4$, then which of the following options is the only CORRECT combination?

• A. $\left(IV\right)\left(iii\right)\left(S\right)$
• B. $\left(IV\right)\left(iv\right)\left(S\right)$
• C. $\left(II\right)\left(iii\right)\left(R\right)$
• D. $\left(II\right)\left(iv\right)\left(R\right)$

Answer 1

The correct option is D $\left(II\right)\left(iv\right)\left(R\right)$

Tangent at $(\sqrt{3},\frac{1}{2})$ is found to be $\sqrt{3}x+2y=4$.
Since, slope of tangent at $(\sqrt{3},\frac{1}{2})$ is negative, hence possible curve are $\left(I\right)$ and $\left(II\right)$ only.
Therefore, equation of curve is
$\frac{x^2}{4}+y^2=1\Rightarrow x^2+a^2{y}^2=a^2.$
Hence, the equation of tangent is $y=mx+\sqrt{a^2{m}^2+1}$.
And the point of contact is $(\frac{-a^{2}m}{\sqrt{a^2{m}^2+1}},\frac{1}{\sqrt{a^2{m}^2+1}})$

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