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# Find the equation of the ellipse if One of the foci is$(0 , -1 )$, the corresponding directrix is $3x+16=0$ and $e=\large\frac{3}{5}$

Toolbox:
• General equation of a conic with focus at $F(x_1,y_1)$,directrix $lx+my+n=0$ and eccentricity 'e' is $(x-x_1)^2+(y-y_1)^2=e^2\bigg[\pm\large\frac{lx+my+n}{\sqrt{l^2+m^2}}\bigg]^2$
Step 1:
$F(0,-1)$ and the corresponding directrix is $3x+16=0$.
$e=\large\frac{3}{5}$
Let $P(x,y)$ be a point on the ellipse and $PM$ be $\perp$ is the directrix.
$\large\frac{FP}{PM}=$$e FP^2=e^2PM^2 Step 2: x^2+(y+1)^2=\large\frac{9}{25}$$\big(\large\frac{\pm 3x+16}{\sqrt 9}\big)^2$
$25(x^2+y^2+2y+1)=9x^2+96x+256$
$16x^2+25y^2-96x+50y-231=0$
This is the required equation.