Question

A tangent to a suitable conic (Column I) 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. $\left(II\right)\left(iv\right)\left(R\right)$
• C. $\left(II\right)\left(iii\right)\left(R\right)$
• D. $\left(IV\right)\left(iv\right)\left(S\right)$

Answer 1

Answer: (B)

$\left(II\right)\left(iv\right)\left(R\right)$

Equation $IV$ is a hyperbola and its tangent equation will be $y=mx+\sqrt{(a^2{m}^2-b^2)}$. Here, $b^2=1$, hence $\left(iii\right)$ in 2nd column matches with $IV$. So option $D$ is discarded.

Equation $II$ is the equation of an ellipse and its tangent equation will be $y=mx+\sqrt{a^2{m}^2+b^2}$. Here, $b^2=1$, hence $\left(iv\right)$ in 2nd column matches with $II$. So, option $C$ is discarded.

For $II$, $x^2+a^2{y}^2=a^2\Rightarrow 3+\frac{a^2}{4}=a^2\Rightarrow 3=\frac{3a^2}{4}\Rightarrow a^2=4$\

For the tangent, $\sqrt{3}x+2y=4$, we get that $m=-\frac{\sqrt{3}}{2}$ and $c=2$.

$c^2=a^2{m}^2+1$

$\Rightarrow a^2{m}^2+1=4\left(\frac{3}{4}\right)+1=3+1=4$

$\Rightarrow c^2=4\Rightarrow c=2$

Hence, option $B$ matches the given criteria and is the correct option.