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

Open in App

Solution

$Let p(h,k) be a point. Let the given points be A(0,2) and B(0,-2) According to the given conditin, A P + B P = 6 \Rightarrow \sqrt{(h - 0)^{2} + (k - 2)^{2}} + \sqrt{(h - 0)^{2} + (k + 2)^{2}} \Rightarrow \sqrt{h^{2} (k - 2) + h^{2}} = 6 - \sqrt{h^{2} + (k + 2)^{2}} Squaring both sides,we get: \Rightarrow h^{2} + (k - 2)^{2} = 36 + h^{2} + (k + 2)^{2} - 12 {\sqrt{h^{2} + (k + 2)}}^{2} \Rightarrow h^{2} + k^{2} + 4 - 4 k = 36 + h^{2} + k^{2} + 4 + 4 k - 12 \sqrt{h^{2} + (k + 2)^{2}} \Rightarrow 3 \sqrt{h^{2} + (k + 2)^{2}} = 9 + 2 k \Rightarrow 9 (h^{2} + k^{2} + 4 + 4 k) = 81 + 4 k^{2} + 36 k (S q u a r i n g b o t h s i d e s) \Rightarrow 9 h^{2} + 9 k^{2} + 36 + 36 k = 81 + 4 k^{2} + 36 k \Rightarrow 9 h^{2} + 5 k^{2} - 45 = 0 H e n c e, t h e l o c u s o f (h, k) i s 9 x^{2} + 5 y^{2} - 45 = 0$

Suggest Corrections

Similar questions

P

(0, 2)

(0, - 2)

6

Find the equation of the locus of a point such that the sum of its distance from (0,2) and (0,-2) is 6

(0, 2)

(0, - 2)

6

(2, 0)

(- 2, 0)

6

Join BYJU'S Learning Program