Find the eccentricity, foci, length of the Latus rectum and the equations of directrices of the ellipse $9 x^{2} + 16 y^{2} = 144$ .

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Solution

Given equation can be written in standard form as $\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1$

$\Rightarrow a^{2} = 16, b^{2} = 9 \Rightarrow a = 4, b = 3$

So eccentricity $, e = \sqrt{1 - \frac{b^{2}}{a^{2}}} = \sqrt{1 - \frac{9}{16}} = \frac{\sqrt{7}}{4}$

Thus foci $= (\pm a e, 0) = (\pm \sqrt{7}, 0)$

Length of latus rectum $= \frac{2 b^{2}}{a} = \frac{2 \cdot 9}{4} = \frac{9}{2}$

And equation of directrices are $x = \pm \frac{a}{e} = \pm \frac{16}{\sqrt{7}}$

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Find the eccentricity, foci, length of the Latus rectum and the equations of directrices of the ellipse 9x2+16y2=144.

Given equation can be written in standard form as x216+y29=1 ⇒a2=16,b2=9⇒a=4,b=3So eccentricity ,e=√1−b2a2=√1−916=√74Thus foci =(±ae,0)=(±√7,0)Length of latus rectum =2b2a=2⋅94=92And equation of directrices are x=±ae=±16√7

Find the eccentricity, foci, length of the Latus rectum and the equations of directrices of the ellipse $9 x^{2} + 16 y^{2} = 144$ .