Question

​​​​​​$\begin{array}{cc}ColumnI& ColumnII\\ \text{a. If}x,y\in R,\text{satisfying the equation}\frac{(x-4{)}^2}{4}+\frac{y^2}{9}=1& \text{p.}-\frac{2}{3}\\ \text{then the difference between the largest and smallest value}& \\ \text{of the expression}\frac{x^2}{4}+\frac{y^2}{9}\text{is}& \\ \text{b. If}PQ\text{is focal chord of ellipse}\frac{x^2}{25}+\frac{y^2}{16}=1\text{which passes}& \text{q.}10\\ \text{through}S\equiv (3,0)\text{and}PS=2,\text{then length of chord}PQ\text{is}& \\ \text{c. If the normal at the point}P\left(\theta \right)\text{to the ellipse}\frac{x^2}{14}+\frac{y^2}{5}=1& \\ \text{intersect it again at the point}Q\left(2\theta \right),\text{then the value of}& \\ \cos \theta \text{is}& \text{r.}\frac{3}{4}\sqrt{7}\\ \text{d. The length of common tangent to}{x}^2+y^2=16\text{and}& \text{s.}8\\ 9x^2+25y^2=225\text{is}& \end{array}$

• A. $a-s;b-r;c-q;d-p$
• B. $a-p;b-q;c-r;d-s$
• C. $a-s;b-q;c-p;d-r$
• D. $a-q;b-r;c-s;d-p$

Answer 1

The correct option is C $a-s;b-q;c-p;d-r$


$\left(a\right).$
Let $x-4=2\cos \theta$,
$x=2\cos \theta +4$ and $y=3\sin \theta$
Now,
$E=\frac{x^2}{4}+\frac{y^2}{9}$
$=\frac{(2cos\theta +4{)}^2}{4}+{\sin }^2\theta$
$=\frac{4{\cos }^2\theta +16+16\cos \theta +4{\sin }^2\theta }{4}$
$=\frac{20+16\cos \theta }{4}$
$=5+4\cos \theta$
Hence,
$E_{max}-E_{min}=9-1=8$

$\left(b\right)$.
We know that
$\Rightarrow \frac{1}{SP}+\frac{1}{SQ}=\frac{2a}{b^2}$
$\Rightarrow \frac{1}{2}+\frac{1}{SQ}=\frac{10}{16}$
$\Rightarrow SQ=8$
$\Rightarrow PQ=10$

$\left(c\right)$.
Let $P\left(\theta \right)=(\sqrt{14}\cos \theta ,\sqrt{5}\sin \theta )$
$\therefore$ Normal at P on the ellipse will be
$\frac{\sqrt{14}x}{\cos \theta }-\frac{\sqrt{8}y}{\sin \theta }=9$
$\because Q\left(2\theta \right)\equiv (\sqrt{14}\cos 2\theta ,\sqrt{5}\sin 2\theta )$ lies on the given eqn of the normal.
$\therefore \frac{\sqrt{14}\left(\sqrt{14}\cos 2\theta \right)}{\cos \theta }-\frac{\sqrt{5}\left(\sqrt{5}\sin 2\theta \right)}{\sin \theta }=9$
$\Rightarrow 28{\cos }^2\theta -14-10{\cos }^2\theta =9\cos \theta$
$\Rightarrow \cos \theta =\frac{7}{6}$, or $-\frac{2}{3}$ $\left(\cos \theta \text{/}>1\right)$
$\Rightarrow \cos \theta =-\frac{2}{3}$

$\left(d\right).$
$9x^2+25y^2=225$
$\Rightarrow \frac{x^2}{25}+\frac{y^2}{9}=1$
$a=5$
$b=3$
Circle $C:x^2+y^2=16$
Eqn of tangent to the ellipse at $A(5\cos \theta ,3\sin \theta )$ is,
$\frac{x\cos \theta }{5}+\frac{y\sin \theta }{3}=1$
Since this line is also a tangent to the circle,
$\therefore \frac{1}{\sqrt{\frac{{\cos }^2\theta }{25}+\frac{{\sin }^2\theta }{9}}}=4$
$\Rightarrow \frac{1}{16}=\frac{{\cos }^2\theta }{25}+\frac{{\sin }^2\theta }{9}$
$\Rightarrow {\sin }^2\theta =\frac{81}{256}\Rightarrow {\cos }^2\theta =\frac{175}{256}$
And we know that
$AB=\sqrt{S_1}$
$=\sqrt{(5\cos \theta {)}^2+(3\sin \theta {)}^2-16}$
$=\sqrt{9+16{\cos }^2\theta -16}$
$=\sqrt{16\times \frac{175}{256}-7}$
$\Rightarrow AB=\frac{3}{4}\sqrt{7}$