Question

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

• A. $\left(I\right)\left(ii\right)\left(Q\right)$
• B. $\left(I\right)\left(i\right)\left(P\right)$
• 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 A $\left(I\right)\left(ii\right)\left(Q\right)$

$a=\sqrt{2}$ and Point of contact : $(-1,1)$

Comparing with 1st column,

$x^2+y^2-a^2=0_{(-1,1)}=1+1-(\sqrt{2}{)}^2=0$

$\therefore$ Equation of curve is $x^2+y^2=a^2$
$\therefore$ Equation of tangent $=2x+2y{y}^{\prime }=0$

$y^{\prime }=-{\frac{x}{y}}_{(-1,1)}=1$

$\Rightarrow \frac{y-1}{x+1}=1$

Hence, $y=x+2$

Comparing with column 2:

1) $my=m^{2}x+a\Rightarrow$ Not possible

2) $y=mx+a\sqrt{m^2+1}$ and $y=x+2$

$m=1$ and $a\sqrt{m^2+1}=2$

$\therefore$ LHS = RHS

Hence, option A is correct.