The equation of the ellipse with directrix 3x | vedclass

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(B) By the definition of an ellipse,the distance from a point $(x, y)$ to the focus is $e$ times the distance from the point to the directrix: $\sqrt{

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$91x^2+84y^2-24xy-170x-360y+475=0$

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(B) By the definition of an ellipse,the distance from a point $(x, y)$ to the focus is $e$ times the distance from the point to the directrix: $\sqrt{(x-1)^2+(y-2)^2} = \frac{1}{2} \frac{|3x+4y-5|}{\sqrt{3^2+4^2}}$.Squaring both sides,we get: $(x-1)^2+(y-2)^2 = \frac{1}{4} \frac{(3x+4y-5)^2}{25}$.$100(x^2-2x+1+y^2-4y+4) = (3x+4y-5)^2$.$100(x^2+y^2-2x-4y+5) = 9x^2+16y^2+25+24xy-30x-40y$.$100x^2+100y^2-200x-400y+500 = 9x^2+16y^2+24xy-30x-40y+25$.Rearranging terms: $91x^2+84y^2-24xy-170x-360y+475=0$.

The equation of the ellipse with directrix $3x+4y-5=0$,focus $(1,2)$ and eccentricity $e = \frac{1}{2}$,is

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