Question

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

For $a=\sqrt{2}$, if a tangent is drawn to a suitable conic $\text{(Column 1)}$ at the point of contact $(-1,1)$, then which of the following options is the only $\text{CORRECT}$ combination for obtaining its equation?

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

Answer 1

The correct option is B $\left(I\right)\left(ii\right)\left(Q\right)$

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

$\because a=\sqrt{2}$ and point $(–1,1)$ satisfies curve $x^2+y^2=a^2$,
$\therefore$ equation of tangent is $y=mx+a\sqrt{m^2+1}$ and point of contact is $(\frac{-ma}{\sqrt{m^2+1}},\frac{a}{\sqrt{m^2+1}})$