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Questions  >>  CBSE XI  >>  Math  >>  Conic Sections
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Q)

Find the equation of the set of all points whose distance from $(0, 4)$ are $\large\frac{2}{3}$ of their distance from the line $y = 9$.

$\begin{array}{1 1}5x^2-5y^2+180=0\\5x^2-5y^2-180=0\\5x^2+5y^2+180=0\\5x^2+5y^2-180=0\end{array} $

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  • Distance between two points $A(x_1,y_1)$ and $B(x_2,y_2)$ is $AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$.
Answer : $5x^2+5y^2+180=0$
Given that the distance of the set of points from $(0,4)$ is $\large\frac{2}{3}$ of the distance from the line $y=9$
Hence $\sqrt{(x-0)^2+(y-4)^2}=\large\frac{2}{3}$$\sqrt{(x-0)^2+(y-9)^2}$
Squaring on both sides we get,
$x^2+(y-4)^2=\large\frac{4}{9}$$\big[(x^2)+(y-9)^2\big]$
$9(x^2+y^2-8y+16)=4x^2+4(y^2-18y+81)$
On simplifying we get,
$5x^2+5y^2+180=0$
This is the required equation of the set of points.
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