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Standard equation of an ellipse - class-XII
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The equation of the ellipse whose equation of directrix is $3x+4y-5=0$, coordinates of the focus are $(1,2)$ and the eccentricity is $\dfrac{1}{2}$ is $91x^2+84y^2-24xy-170x-360y+475=0$
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A
True
💡 Explanation:
Let $P(x,y)$ be any point on the ellipse and PM be the perpendicular from P upon the directrix $3x+4y-5=0$.
Then by the definition,
$\dfrac{SP}{PM}=e$
$SP=e.PM$
$\sqrt{(x-1)^2+(y-2)^2}=\dfrac{1}{2}|\dfrac{3x+4y-5}{\sqrt{3^2+4^2}}|$
$(x-1)^2+(y-2)^2=\dfrac{1}{4}. \dfrac{(3x+4y-5)^2}{25}$
$100(x^2+y^2-2x-4y+5)=9x^2+16y^2+24xy-30x-40y+25$
$91x^2+84y^2-24xy-170x-360y+475=0$ is the equation of the ellipse.