Find the eccentricity, centre, foci, vertices of the ellipse $16 x^{2} + 9 y^{2} + 32 x - 36 y = 92$ and also draw the diagram.

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Solution

$16 x^{2} + 9 y^{2} + 32 x - 36 y = 92$
$16 x^{2} + 32 x + 9 y^{2} - 36 y = 92$
$16 (x^{2} + 2 x + 1) - 16 + 9 (y^{2} - 4 y + 4) - 36 = 92$
$\Rightarrow 16 {(x + 1)}^{2} + 9 {(y - 2)}^{2} = 92 + 36 + 16$
$= 144$
$\Rightarrow \frac{{(x + 1)}^{2}}{9} + \frac{{(y - 2)}^{2}}{16} = 1$
So, center $= (- 1, 2), 9 = 16 (1 - e^{2})$
$\Rightarrow 1 - \frac{9}{16} = e^{2}$
$\Rightarrow e = \frac{\sqrt{7}}{4}$
$\Rightarrow a e = \sqrt{7}$
So, focii are $= (- 1, 2 + \sqrt{7}), (- 1, 2 - \sqrt{7})$
Vertices are $(2, 2), (- 4, 2), (- 1, - 2), (- 1, 6)$

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Find the eccentricity, centre, foci, vertices of the ellipse 16x2+9y2+32x−36y=92 and also draw the diagram.

16x2+9y2+32x−36y=9216x2+32x+9y2−36y=9216(x2+2x+1)−16+9(y2−4y+4)−36=92⇒16(x+1)2+9(y−2)2=92+36+16=144⇒(x+1)29+(y−2)216=1So, center =(−1,2),9=16(1−e2)⇒1−916=e2⇒e=√74⇒ae=√7So, focii are =(−1,2+√7),(−1,2−√7)Vertices are (2,2),(−4,2),(−1,−2),(−1,6)

Find the eccentricity, centre, foci, vertices of the ellipse $16 x^{2} + 9 y^{2} + 32 x - 36 y = 92$ and also draw the diagram.