Question

Match Elements of List I with elements of List II.
List I List II

A. If the tangent to the ellipse $x^2+4y^2=16$ at the point $P\left(\varphi \right)$ is a normal to the circle $x^2+y^2-8x-4y=0$, then $\frac{\varphi }{2}$ may be	1. $0$
B. The eccentric angle(s) of a point on the ellipse $x^2+3y^2=6$ at a distance $2$ units from the centre of the ellipse is/are	2. ${\cos }^{-1}(-\frac{2}{3})$
C. The eccentric angle of intersection of the ellipse $x^2+4y^2=4$ and the parabola $x^2+1=y$ is	3. $\frac{\pi }{4}$
D. If the normal at the point $P\left(\theta \right)$ to the ellipse $\frac{x^2}{14}+\frac{y^2}{5}=1$ intersects it again at the point $Q\left(2\theta \right)$, then $\theta$ is	4. $\frac{5\pi }{4}$

• A. A-4,3 , B-2,4, C-3, D-1
• B. A-2,3 , B-3, C-4, D-1
• C. A-1,3 , B-3,4, C-1, D-2
• D. A-1,4 , B-1,2, C-3, D-2

Answer 1

Answer: (A)

A-4,3 , B-2,4, C-3, D-1

A:
Tangent to ellipse at $P\left(\varphi \right)$ is $\frac{x}{4}\cos \varphi +\frac{y}{2}\sin \varphi =1$.
It must pass through the centre of the circle. Hence,
$\frac{4}{4}\cos \varphi +\frac{2}{2}\sin \varphi =1$
$\Rightarrow \cos \varphi +\sin \varphi =1$$\sqrt{1-{\sin }^2\varphi }+\sin \varphi =1$$1-si{n}^2\varphi =(1-\sin \varphi {)}^2$$\sin \varphi -{\sin }^2\varphi =0$
$\Rightarrow 1+\sin 2\varphi =1$
$\Rightarrow \varphi =0$ or $\frac{\pi }{2}$
$\Rightarrow \frac{\varphi }{2}=0$ or $\frac{\pi }{4}$
B:
Consider any point P $(\sqrt{6}\cos \theta ,\sqrt{2}\sin \theta )$ on ellipse $\frac{x^2}{6}+\frac{y^2}{2}=1$
Given that $OP=2$
$\Rightarrow 6{\cos }^2\theta +2{\sin }^2\theta =4$
$\Rightarrow 4{\cos }^2\theta =2$
$\Rightarrow \cos \theta =\pm \frac{1}{\sqrt{2}}$
$\Rightarrow \theta =\frac{\pi }{4}$ or $\frac{5\pi }{4}$
C:
Solving the equation of ellipse and parabola (eliminating $x^2$), we have
$\Rightarrow y-1+4y^2=4$
$\Rightarrow 4y^2+y-5=0$
$\Rightarrow 4y^2+y-5=0$
$\Rightarrow (4y+5)(y-1)=0$
$\to y=1,x=0$
The curves touch at $(0,1)$. So, the angle of intersection is $0$.
D:
The normal at $P(a\cos \theta ,\sin \theta )$ is
$\frac{ax}{\cos \theta }-\frac{bx}{\sin \theta }=a^2-b^2$
where $a^2=14,b^2=5$.
It meets the curve again at $Q\left(2\theta \right)$, i.e. $(a\cos 2\theta ,b\sin 2\theta )$. Hence, $\frac{a}{\cos \theta }a\cos 2\theta -\frac{b}{\sin \theta }\left(b\sin 2\theta \right)=a^2-b^2$
$\Rightarrow \frac{14}{\cos \theta }\cos 2\theta -\frac{5}{\sin \theta }\left(\sin 2\theta \right)=14-5$
$\Rightarrow 28{\cos }^2\theta -14-10{\cos }^2\theta =9\cos \theta$
$\Rightarrow 18{\cos }^2\theta -9\cos \theta -14=0$
$\Rightarrow (6\cos \theta -7)(3\cos \theta -2)=0$
$\Rightarrow \cos \theta =-\frac{2}{3}$