Question

$$\begin{array}{cc}\text{Column A}& \text{Column B}\\ \text{(A) The angle between asymptotes of the hyperbola}& \\ 5x^2-2xy-y^2+4x+6y+1=0\text{is}{\tan }^{-1}\left(\sqrt{\frac{m}{n}}\right)& \\ \text{(where}m,n\text{are coprime)}\text{, then}m+n\text{is}& \\ & \text{(P)}2\\ \text{(B) Distance between the focii of the curve represented}& \\ \text{by the equation}x=2+5\cos \theta \text{and}y=3+4\sin \theta \text{is}& \text{(Q)}5\\ \text{(C) The number of distinct normal possible from}& \\ (\frac{11}{4},\frac{1}{4})\text{to the parabola}{y}^2=4x\text{is}& \text{(R)}6\\ \text{(D) The pair of straight lines represented by}& \\ x^2+y^2+3xy+4x+y-1=0\text{intersect at}P.& \\ \text{If}Q\text{and}R\text{are the points of intersection of the}& \\ \text{pair of lines with the x-axis and the area of the}& \\ \Delta PQR\text{is}A,\text{then}\frac{A^2}{4}\text{is}& \\ & \text{(S)}7\\ & \text{(T)}8\end{array}$$

Which of the following is the only CORRECT combination?

• A. $\left(A\right)\to \left(T\right)$
• B. $\left(B\right)\to \left(R\right)$
• C. $\left(C\right)\to \left(Q\right)$
• D. $\left(D\right)\to \left(S\right)$

Answer 1

The correct option is B $\left(B\right)\to \left(R\right)$

$\left(A\right)$
The combined equation of asymptotes of the hyperbola differs from the equation of asymptotes.
$5x^2-2xy-y^2+4x+6y+1=0$
$a=5,h=-1,b=-1$
$\tan \theta =\left|\frac{2\sqrt{h^2-ab}}{a+b}\right|=\left|\frac{2\sqrt{6}}{4}\right|=\sqrt{\frac{3}{2}}$
$\Rightarrow m+n=5$
$\left(A\right)\to \left(Q\right)$

$\left(B\right)$
For $x=2+5\cos \theta$ and $y=3+4\sin \theta$,
$\frac{(x-2{)}^2}{25}+\frac{(y-3{)}^2}{16}=0$
$e^2=1-\frac{b^2}{a^2}=\frac{9}{25}$
Focii distance $=2ae=6$
$\left(B\right)\to \left(R\right)$

$\left(C\right)$
For the parabola $y^2=4x$,
The equation of the normal to the parabola is,
$y+tx=2t+t^3$ at $(t^2,2t)$
It passes through $(\frac{11}{4},\frac{1}{4})$
$\Rightarrow \frac{1}{4}=-\frac{11}{4}t+2t+t^3$
$\Rightarrow (t-1)(2t+1{)}^2=0$
Therefore, two distinct normals are possible.
$\left(C\right)\to \left(P\right)$

$\left(D\right)$
For $x^2+y^2+3xy+4x+y-1=0$
differentiating with respect to $x$,
$\Rightarrow 2x+3y+4=0\cdots \left(1\right)$
differentiating with respect to $y$,
$\Rightarrow 3x+2y+1=0\cdots \left(2\right)$

Solving $\left(1\right)$ and $\left(2\right)$, we get
$P(1,-2)$
These lines cut x-axis at $Q$ and $R$
$x^2+4x-1=0\Rightarrow (x+2{)}^2=5\Rightarrow x=-2\pm \sqrt{5}$
$QR=2\sqrt{5}$
Area
$A=\frac{1}{2}\times 2\sqrt{5}\times 2=2\sqrt{5}$
$\Rightarrow \frac{A^2}{4}=5$
$\left(D\right)\to \left(Q\right)$