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# Find the equation of ellipse whose eccentricity is 2/3 ,

Question:

Find the equation of ellipse whose eccentricity is 2/3 , latus rectum is 5 and the centre is (0, 0).

Solution:

We know that equation of an ellipse

$=\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$

Also we have,

Length of latus rectum $=\frac{2 b^{2}}{a}$

Length of minor $a x i s=2 b$

Given

$e=\frac{2}{3}$

$\mathrm{b}^{2}=\mathrm{a}^{2}\left(1-\mathrm{e}^{2}\right) \ldots \ldots \ldots 1$

Length of Latus Rectum $=\frac{2 \mathrm{~b}^{2}}{\mathrm{a}}=5$

$\mathrm{b}^{2}=2.5 \mathrm{a} \ldots \ldots \ldots \ldots .2$

Now by substituting the values in equation 1 we get

$\therefore 2.5 \mathrm{a}=\mathrm{a}^{2}\left(1-\left(\frac{2}{3}\right)^{2}\right)$

On simplifying and rearranging we get

$2.5 a=a^{2}\left(1-\frac{4}{9}\right)$

$2.5 a=a^{2}\left(\frac{5}{9}\right)$

$22.5 \mathrm{a}=5 \mathrm{a}^{2}$

$5 a^{2}-22.5 a=0$

$5 a(a-4.5)=0$

$a=0\{$ Not Possible, as the length of latus rectum is 5 units $\}$ or $4.5=9 / 2$ Now by substituting the values in equation 2 we get

$\mathrm{b}^{2}=2.5 \times 4.5=11.25=\frac{45}{4}$

Again we have equation of an ellipse

$=\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$

By substituting this values we get

$\frac{\mathrm{x}^{2}}{\left(\frac{9}{2}\right)^{2}}+\frac{\mathrm{y}^{2}}{\frac{45}{4}}=1$

$\frac{4 x^{2}}{81}+\frac{4 y^{2}}{45}=1$

$\therefore$ Equation of an ellipse is $\frac{4 x^{2}}{81}+\frac{4 y^{2}}{45}=1$