For the ellipse given by frac{(x-3)^2}{25}+fr | vedclass

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(A) The given equation of the ellipse is $\frac{(x-3)^2}{25}+\frac{(y-2)^2}{16}=1$. Comparing this with the standard form $\frac{(x-h)^2}{a^2}+\frac{(

Vedclass · Accepted Answer

Vedclass · Answer

(A) The given equation of the ellipse is $\frac{(x-3)^2}{25}+\frac{(y-2)^2}{16}=1$. Comparing this with the standard form $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$,we have $h=3, k=2, a^2=25, b^2=16$. Thus,$a=5$ and $b=4$. The eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$. $(i)$ The equation of the major axis is $y = k$,so $y = 2$. This matches $(q)$. $(ii)$ The equations of the directrices are $x = h \pm \frac{a}{e}$. $x = 3 \pm \frac{5}{3/5} = 3 \pm \frac{25}{3}$. $x = 3 + \frac{25}{3} = \frac{34}{3} \Rightarrow 3x = 34$. $x = 3 - \frac{25}{3} = -\frac{16}{3} \Rightarrow 3x = -16$. Thus,$3x = 34$ matches $(p)$. $(iii)$ The equations of the latera recta are $x = h \pm ae$. $x = 3 \pm (5 	imes \frac{3}{5}) = 3 \pm 3$. $x = 3 + 3 = 6$ or $x = 3 - 3 = 0$. Thus,$x = 6$ matches $(s)$. Therefore,the correct matches are $(i) ightarrow (q)$,$(ii) ightarrow (p)$,$(iii) ightarrow (s)$.

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 latus rectum	$(r)$ $x + y = 9$
	$(s)$ $x = 6$
	$(t)$ $x = 3$
	$(u)$ $3y = 34$

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