Question

$\begin{array}{cccccc}& \text{Column I}& & \text{Column II}& & \text{Column III}\\ \left(1\right)& x^2+y^2=a^2& \left(A\right)& y=mx+\frac{a}{m}& \left(P\right)& x^2+y^2=2a^2\\ \left(2\right)& y^2=4ax& \left(B\right)& y=mx\pm \sqrt{a^2{m}^2+b^2}& \left(Q\right)& x^2+y^2=a^2+b^2\\ \left(3\right)& \frac{x^2}{a^2}+\frac{y^2}{b^2}=1& \left(C\right)& y=mx\pm \sqrt{a^2{m}^2-b^2}& \left(R\right)& x^2+y^2=a^2-b^2\\ \left(4\right)& \frac{x^2}{a^2}-\frac{y^2}{b^2}=1& \left(D\right)& y=mx\pm a\sqrt{1+m^2}& \left(S\right)& x=-a\end{array}$

$a)$ Column $\text{I}$ represents the equation of the conic.
$b)$ Column $\text{II}$ represents the slope form of the tangent.
$a)$ Column $\text{III}$ represents the director circle of the conic from column $\text{I}$

Which of the following is only correct combination?

• A. $\left(1\right)\to \left(A\right)\to \left(P\right)$
• B. $\left(2\right)\to \left(A\right)\to \left(P\right)$
• C. $\left(3\right)\to \left(B\right)\to \left(Q\right)$
• D. $\left(4\right)\to \left(C\right)\to \left(Q\right)$

Answer 1

The correct option is C $\left(3\right)\to \left(B\right)\to \left(Q\right)$

$\begin{array}{ccc}\text{Curve}& \text{Tangent}& \text{Director circle}\\ x^2+y^2=a^2& y=mx\pm a\sqrt{1+m^2}& x^2+y^2=2a^2\\ y^2=4ax& y=mx+\frac{a}{m}& x=-a\\ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1& y=mx\pm \sqrt{a^2{m}^2+b^2}& x^2+y^2=a^2+b^2\\ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1& y=mx\pm \sqrt{a^2{m}^2-b^2}& x^2+y^2=a^2-b^2\end{array}$