Question

Find the locus of a point such that the sum of its distances from (0, 2) and (0, −2) is 6.

 

Answer 1

Let P(h, k) be a point. Let the given points be \[A\left( 0, 2 \right)\text{ and B}\left( 0, - 2 \right)\]
According to the given condition,
AP + BP = 6
⇒ \[\sqrt{\left( h - 0 \right)^2 + \left( k - 2 \right)^2} + \sqrt{\left( h - 0 \right)^2 + \left( k + 2 \right)^2} = 6\]

⇒ \[\sqrt{h^2 + \left( k - 2 \right)^2} = 6 - \sqrt{h^2 + \left( k + 2 \right)^2}\]

Squaring both sides, we get:
⇒ \[h^2 + \left( k - 2 \right)^2 = 36 + h^2 + \left( k + 2 \right)^2 - 12\sqrt{h^2 + \left( k + 2 \right)^2}\]

⇒ \[h^2 + k^2 + 4 - 4k = 36 + h^2 + k^2 + 4 + 4k - 12\sqrt{h^2 + \left( k + 2 \right)^2}\]

⇒ \[3\sqrt{h^2 + \left( k + 2 \right)^2} = 9 + 2k\]
⇒ \[9\left( h^2 + k^2 + 4 + 4k \right) = 81 + 4 k^2 + 36k\]    (Squaring both sides)

\[\Rightarrow 9 h^2 + 9 k^2 + 36 + 36k = 81 + 4 k^2 + 36k\]

⇒ \[9 h^2 + 5 k^2 - 45 = 0\]
Hence, the locus of (h, k) is \[9 x^2 + 5 y^2 - 45 = 0\].