Question

For the ellipse given by $\frac{(x-3{)}^2}{25}+\frac{(y-2{)}^2}{16}=1$, match the equations of the lines given in List I with those on the List II.

	List I		List II
(i)	The equation of the major axis	(p)	$3x=34$
(ii)	The equation of a directrix	(q)	$y=2$
(iii)	The equation of a latusrectum	(r)	$x+y=9$
		(s)	$x=6$
		(t)	$x=3$
		(u)	$3y=34$

• A. (i) - (t) ,(ii) - (p) ,(iii) - (s)
• B. (i) - (q) ,(ii) - (u) ,(iii) - (t)
• C. (i) - (q) ,(ii) - (p) ,(iii) - (t)
• D. (i) - (q) ,(ii) - (p) ,(iii) - (s)

Answer 1

Answer: (D)

Given equation of ellipse
$\frac{(x-3{)}^2}{25}+\frac{(y-2{)}^2}{16}=1$
Let $(x-3)=X$ and $(y-2)=Y$, then
$\frac{x^2}{5^2}+\frac{y^2}{4^2}=1$
Here, $a=5$ and $b=4$.
Also, $e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{16}{25}}=\frac{3}{5}$
Now, equation of the major axis $Y=0$
$\Rightarrow y-2=0$
$\therefore y=2$
So, $(i)\to (q)$
Equation of a directrix
$X=\frac{a}{e}=\frac{5}{3}\times 5=\frac{25}{3}$
$\Rightarrow x-3=\frac{25}{3}$
$\Rightarrow x=\frac{25}{3}+3=\frac{34}{3}$
$\Rightarrow 3x=34$
So, $(ii)\to (p)$
Equation of a latusrectum,
$X=\pm$ ae
$\Rightarrow x-3=\pm ae$
$\Rightarrow x-3=\pm 5\times \frac{3}{5}$
$\Rightarrow x=\pm 3+3$
$\therefore x=6,0$
So, $(iii)\to (s)$