Question
$16 x^2+25 y^2=400$
✓
Answer
Given equation of the ellipse is $16 x^2+25 y^2=400$
$\frac{x^2}{25}+\frac{y^2}{16}=1$
Comparing this equation with $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, we get
$a^2=25 \text { and } b^2=16$
$\therefore a=5 \text { and } b=4$
$\text { Since } a>b,$
X-axis is the major axis and Y-axis is the minor axis
(i) Length of major axis = 2a = 2(5) = 10
Length of minor axis = 2b = 2(4) = 8
Lengths of the principal axes are 10 and 8.
$b^2=a^2\left(1-e^2\right)$
$16=25\left(1- e ^2\right)$
$\frac{16}{25}=1- e ^2$
$e ^2=1-\frac{16}{25}$
$e ^2=\frac{9}{25}$
$e =\frac{3}{5} \ldots \ldots[\because 0< e <1]$
Co-ordinates of the foci are S(ae, 0) and S'(-ae, 0),
i.e., $S\left(5\left(\frac{3}{5}\right), 0\right)$ and $S^{\prime}\left(-5\left(\frac{3}{5}\right), 0\right)$,
i.e., S(3, 0) and S'(-3, 0)
Equations of the directrices are $x= \pm \frac{a}{e}$
i.e., $x= \pm \frac{5}{\left(\frac{3}{5}\right)}$
i.e., $x= \pm \frac{25}{3}$
(iv) Length of latus rectum $=\frac{2 b^2}{a}=\frac{2(16)}{5}=\frac{32}{5}$
(v) Distance between foci $=2 ae =2(5)\left(\frac{3}{5}\right)=6$
(vi) Distance between directrices $=\frac{2 a}{e}=\frac{2(5)}{\left(\frac{3}{5}\right)}=\frac{50}{3}$
$\frac{x^2}{25}+\frac{y^2}{16}=1$
Comparing this equation with $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, we get
$a^2=25 \text { and } b^2=16$
$\therefore a=5 \text { and } b=4$
$\text { Since } a>b,$
X-axis is the major axis and Y-axis is the minor axis
(i) Length of major axis = 2a = 2(5) = 10
Length of minor axis = 2b = 2(4) = 8
Lengths of the principal axes are 10 and 8.
$b^2=a^2\left(1-e^2\right)$
$16=25\left(1- e ^2\right)$
$\frac{16}{25}=1- e ^2$
$e ^2=1-\frac{16}{25}$
$e ^2=\frac{9}{25}$
$e =\frac{3}{5} \ldots \ldots[\because 0< e <1]$
Co-ordinates of the foci are S(ae, 0) and S'(-ae, 0),
i.e., $S\left(5\left(\frac{3}{5}\right), 0\right)$ and $S^{\prime}\left(-5\left(\frac{3}{5}\right), 0\right)$,
i.e., S(3, 0) and S'(-3, 0)
Equations of the directrices are $x= \pm \frac{a}{e}$
i.e., $x= \pm \frac{5}{\left(\frac{3}{5}\right)}$
i.e., $x= \pm \frac{25}{3}$
(iv) Length of latus rectum $=\frac{2 b^2}{a}=\frac{2(16)}{5}=\frac{32}{5}$
(v) Distance between foci $=2 ae =2(5)\left(\frac{3}{5}\right)=6$
(vi) Distance between directrices $=\frac{2 a}{e}=\frac{2(5)}{\left(\frac{3}{5}\right)}=\frac{50}{3}$
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