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

Find the equation of the set of all points the sum of whose distances from the points $(3, 0)$ and $(9, 0)$ is 12.

$\begin{array}{1 1}3x^2+4y^2+36x=0\\3x^2-4y^2+36x=0\\3x^2+4y^2-36x=0\\3x^2-4y^2-36x=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 : $3x^2+4y^2-36x=0$
Let $S(3,0)$ and $S_1(9,0)$ be the two foci and $P(x,y)$ be the moving point.
It is given that the sum of the distances from the point $(3,0)$ and $(9,0)$ is $12$.
$\therefore \sqrt{(x-9)^2+y^2}+\sqrt{(x-3)^2+y^2}=12$
(ie) $\sqrt{(x-9)^2+y^2}=12-\sqrt{(x-3)^2+y^2}$
Squaring on both sides we get,
$(x-9)^2+y^2=144+(x-3)^2+y^2-24\sqrt{(x-3)^2+y^2}$
On simplifying we get,
$12x+72=24\sqrt{(x-3)^2+y^2}$
(ie) $x+6=2\sqrt{(x-3)^2+y^2}$
Squaring once again we get,
$(x+6)^2=4[(x-3)^2+y^2]$
On simplifying we get,
$3x^2+4y^2-36x=0$
This is the required equation of the set of points.
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