Question

Column-1 contains a conic ' $C$ '
Column-2 contains length of latus rectum of conic ' $C$ '
Column- 3 contains the least distance between the conic ' $C$ ' and its director circle

Column I		Column II		Column III
I	$\sqrt{2(x-1{)}^2+2(y+1{)}^2}=2\left|\frac{3x-4y+1}{5}\right|$	i	$\sqrt{2}$	P	$\frac{1}{5}$
II	$5y^2-10y-4x+1=0$	ii	$\frac{16\sqrt{2}}{5}$	Q	$\sqrt{2}-1$
III	$\sqrt{(x-1{)}^2+(y+1{)}^2}=\left|\frac{3x-4y+1}{25}\right|$	iii	$\frac{16}{25}$	R	$\frac{8\sqrt{2}}{5}$
IV	$x^2-2y^2-2x-4y-3=0$	iv	$\frac{4}{5}$	S	$\frac{2}{15}$

Which of the following is a CORRECT combination ?

• A. (I) (iii) (S)
• B. (II) (iv) (P)
• C. (III) (ii) (S)
• D. (IV) (iv) (Q)

Answer 1

Answer: (B)

(I)$\sqrt{2(x-1{)}^2+2(y+1{)}^2}=2\left|\frac{3x-4y+1}{5}\right|$
$\Rightarrow \sqrt{(x-1{)}^2+(y+1{)}^2}=\sqrt{2}\left|\frac{3x-4y+1}{5}\right|$
Focus $(1,-1)$ and directrix $3x-4y+1=0$
It represent hyperbola with $e=\sqrt{2}$
ae $-\frac{a}{e}=$ perpendicular distance from focus to directrix
$a\left(\frac{2-1}{\sqrt{2}}\right)=\frac{8}{5}\Rightarrow a=\frac{8\sqrt{2}}{5}$
$\Theta b^2=a^2\left(e^2-1\right)$

Length of latus rectum $=\frac{2b^2}{a}=\frac{2a^2\left(e^2-1\right)}{a}=2\times \frac{8\sqrt{2}}{5}(2-1)=\frac{16\sqrt{2}}{5}$
Least distance between the conic and its director circle is $\left|\sqrt{a^2-b^2}-a\right|$
$=\left|\sqrt{a^2-a^2\left(e^2-1\right)}-a\right|=\left|a\sqrt{2e^2}-a\right|=\frac{8\sqrt{2}}{5}.$
(II)
$5y^2-10y-4x+1=0\Rightarrow 5\left(y^2-2y+1\right)=4x+4$
$\Rightarrow 5(y-1{)}^2=4(x+1)\Rightarrow (y-1{)}^2=\frac{4}{5}(x+1)$
It represent parabola.
Focus $(-1,1)$, directrix $x+1=\frac{-1}{5}$ (director circle)
$\Rightarrow x=\frac{-6}{5}$
Minimum distance between conic and director circle is $=\left|\frac{-6}{5}-(-1)\right|=\frac{1}{5}$.
(III) $\sqrt{(x-1{)}^2+(y+1{)}^2}=\frac{1}{5}\left|\frac{3x-4y+1}{5}\right|$
Focus $(1,-1)$; directrix $=3x-4y+1$
It represent ellipse with $e=\frac{1}{5}$
$\Theta \frac{a}{e}-ae=$ perpendicular distance from locus to directrix
$a\left(\frac{1-e^2}{e}\right)=\frac{8}{5}\Rightarrow a=\frac{1}{3}$
$\Theta e^2=1-\frac{b^2}{a^2}\Rightarrow b^2=a^2\left(1-e^2\right)$.
Length of latus rectum $=\frac{2b^2}{a}=\frac{2a^2\left(1-e^2\right)}{a}=\frac{16}{25}$
least distance between conic and director circle is $\left|\sqrt{a^2+b^2}-a\right|=\left|a\sqrt{2-e^2}-a\right|$
$\left|\frac{1}{3}\sqrt{2-\frac{1}{25}}-\frac{1}{3}\right|=\left|\frac{1}{3}\left(\frac{7}{5}-1\right)\right|=\frac{2}{15}.$
(IV)$x^2-2y^2-2x-4y-3=0\Rightarrow \left(x^2-2x+1\right)-\left(2y^2+4y+2\right)=2$
$\Rightarrow (x-1{)}^2-2(y+1{)}^2=2\Rightarrow \frac{(x-1{)}^2}{2}-\frac{(y+1{)}^2}{1}=1$
focus $(1,-1)$
It represents hyperbola.
Length of latus rectum $=\frac{2b^2}{a}=\frac{2(1{)}^2}{\sqrt{2}}=\sqrt{2}$
Least distance between conic and director circle is $\left|\sqrt{a^2-b^2}-a\right|$
$\Rightarrow |\sqrt{2-1}-\sqrt{2}|=\sqrt{2}-1$