Question

foci(-1,-3) ,directrix x-2y=0 e=4/5 find the equation of ellipse

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Answer 1

Given:

• Foci=(-1,-3)

•Directrix >> x-2y=0

• Eccentricity (e) =4/5

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To Find :

• The equation of ellipse =?

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Solution :

➡ Let S(-1,-3) be the focus and ZZ' be the directrix.

Let P(x,y) be any point on the ellipse.

Let PM be the perpendicular from P on the Directrix.

Then by the definition of ellipse , we have :

$\\ \implies\sf{SP=e.PM \: \: \: where \: e=\dfrac{1}{2}}$

On squaring both sides,

$\\ \implies\sf{ SP^{2}=e^{2}.PM^{2}}$

$\\ \implies \sf \: {(x + 1)}^{2} + {(y + 3)}^{2} = \bigg( { \frac{1}{2} } \bigg)^{2} \times \bigg( { \frac{x -2y }{ \sqrt{ {(1)}^{2} + {( - 2)}^{2} } } } \bigg)^{2}$

$\\ \implies\sf{(x+1)^{2}+(y+3)^{2}=\dfrac{1}{4} \bigg|\dfrac{x^{2}+4y^{2}-4xy}{\sqrt{5}}\bigg|}$

$\\ \implies\sf{x^{2}+1+2x+y^{2}+6y+9=\dfrac{1}{4}\bigg|\dfrac{x^{2}+4y^{2}-4xy}{\sqrt{5}}\bigg|}$

$\\ \implies\sf{4x^{2}+4y^{2}+8x+24y+40=\dfrac{x^{2}+4y^{2}-4xy}{\sqrt{5}}}$

$\\ \implies\sf{4\sqrt{5}x^{2}+4\sqrt{5}y^{2}+8\sqrt{5}x+24\sqrt{5}y+40\sqrt{5}-x^{2}-4y^{2}+4xy=0}$

$\\ \implies\sf{(4\sqrt{5}-1)x^{2}+4(\sqrt{5}-1)y^{2}+4xy+8\sqrt{5}x+24\sqrt{5}y=0}$

• This is the equation of the required ellipse.