Question

Find the equation of the locus of a point the sum of whose distances from (0,2)and (0,-2) is 6​

Answer 1

Let us assume that the required point of the locus is ( h, k ) .

According to the distance Formula -

Distance Between Two Points -

$\sf{ \sqrt{ ( x_2 - x_1 ) ^ 2 + ( y_2 - y_1 ) ^ 2 } }$

Let us assume that Point A ( 0, 2 )

Distance between A and Locus -

$\sf{ \sqrt{ h^2 + ( k - 2 ) ^ 2 }}$

Let us assume that Point B ( 0, -2 )

Distance between B and Locus -

$\sf{ \sqrt{ h^2 + ( k + 2 ) ^ 2 }}$

Sum = 6

$\sf{ \sqrt{ h^2 + ( k - 2 ) ^ 2 } + \sqrt{ h^2 + ( k + 2 ) ^ 2 } = 36} \\ \\ \sf{ Squaring \: Both \: Sides \: - } \\ \\ \sf{ 2h^2 + 2k^2 + 8 + \sqrt{ h^4 + h^2 { [ k^2 + 4 ] } + k^4 - 4k + 16 } - 36 = 0 }$

Simplifying that , we get the required equation .

Now, replace h and k as x and y respectively to get the answer .

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