Find the locus of the point P such that the sum of it's distances from (0,2) and (0,-2) is 6 - Maths - Straight Lines

Question

Find the locus of the point P such that the sum of it's
distances from (0,2) and (0,-2) is 6

Chetna Khanna · Accepted Answer

Dear Student Let P(x, y) be any point on the locus Given, PA + PB = 6, where coordinates of point A is (0, 2) and B is (0, â€“ 2) or PA = 6 â€“ PB $\Rightarrow \sqrt{{(x - 0)}^{2} + (y - 2)^{2}} = 6 - \sqrt{{(x - 0)}^{2} + (y + 2)^{2}}$ Squaring both the sides $x^{2} + {(y - 2)}^{2} = 6^{2} + {x^{2} + (y + 2)^{2}} - 12 \sqrt{x^{2} + (y + 2)^{2}} \Rightarrow x^{2} + y^{2} + 4 - 4 y = 36 + x^{2} + y^{2} + 4 + 4 y - 12 \sqrt{x^{2} + y^{2} + 4 + 4 y} \Rightarrow - 8 y - 36 = - 12 \sqrt{x^{2} + y^{2} + 4 + 4 y} \Rightarrow - 4 (2 y + 9) = - 12 \sqrt{x^{2} + y^{2} + 4 + 4 y} \Rightarrow 2 y + 9 = 3 \sqrt{x^{2} + y^{2} + 4 + 4 y}$ Squaring both the sides ${(2 y + 9)}^{2} = 3^{2} (x^{2} + y^{2} + 4 + 4 y) \Rightarrow 4 y^{2} + 81 + 36 y = 9 x^{2} + 9 y^{2} + 36 + 36 y \Rightarrow 9 x^{2} + 5 y^{2} - 45 = 0$ So, the locus of the point P is 9x2 + 5y2 â€“ 45 = 0 Cheers!!

Find the locus of the point P such that the sum of it's distances from (0,2) and (0,-2) is 6.