Question

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

Answer 1

$\text{Let P(h,k) be the moving point}$

$\text{Let the given points be A(0,2) and B(0,-2)}$

$\textbf{Given: }PA+PB=6$

$\sqrt{(h-0)^2+(k-2)^2}+\sqrt{(h-0)^2+(k+2)^2}=6$

$\sqrt{h^2+(k-2)^2}=6-\sqrt{h^2+(k+2)^2}$

$\text{squaring on both sides, we get}$

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

$h^2+k^2+4-4k=36+(h^2+k^2+4+4k)-12\sqrt{h^2+(k+2)^2}$

$-4k=36+4k-12\sqrt{h^2+(k+2)^2}$

$-8k-36=-12\sqrt{h^2+(k+2)^2}$

$-4(2k+9)=-12\sqrt{h^2+(k+2)^2}$

$2k+9=3\sqrt{h^2+(k+2)^2}$

$\text{squaring once again on both sides, we get}$

$(2k+9)^2=9[h^2+(k+2)^2]$

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

$4k^2+81+36k=9h^2+9k^2+36+36k]$

$4k^2+81=9h^2+9k^2+36$

$9h^2+5k^2=45$

$\displaystyle\frac{h^2}{5}+\frac{k^2}{9}=1$

$\therefore\textsf{The locus of P is}$

$\displaystyle\mathsf{\frac{x^2}{5}+\frac{y^2}{9}=1}$

Find more:

B and c are fixed points having coordinates (3,0) and (-3,0) respectively. If the vertical angle bac is 90 then locus

https://brainly.in/question/13799728#

Answer 2

AnswEr:

Let P (h,k) be any point on the locus and let A (0,2) and B (0,-2) be the given points. By the given condition.

PA + PB = 6

$\hookrightarrow \sf \: \sqrt{ {(h - 0)}^{2} + {(k + 2)}^{2} } + \sqrt{ {(h - 0) + {(k + 2)}^{2} } } \\ \\ \hookrightarrow \sf \: \sqrt{ {h}^{2} + {(k - 2)}^{2} } = 6 - \sqrt{ {h}^{2} - {(k + 2)}^{2} } \\ \\ \hookrightarrow \sf \: {h}^{2} + {(k - 2)}^{2} = 36 - 12 \sqrt{ {h}^{2} + {(k + 2)}^{2} } + \\ \sf {h}^{2} + {(k + 2)}^{2} \\ \\ \hookrightarrow \sf \: - 8k - 36 = - 12 \sqrt{ {h}^{2} + {(k + 2)}^{2} } \\ \\ \hookrightarrow \sf \:(2k + 9) = 3 \sqrt{ {h}^{2} + {(k + 2)}^{2} } \\ \\ \hookrightarrow \sf \: {(2k + 9)}^{2} = 9( {h}^{2} + {(k + 2)}^{2} \\ \\ \hookrightarrow \sf \:4 {k}^{2} + 36k + 81 \\ \sf = 9 {h}^{2} + 9 {k}^{2} + 36k + 36 \\ \\ \hookrightarrow \sf \: 9{h}^{2} + 5 {k}^{2} = 45$

Hence, locus of (h,k) is 9x² + 5y² = 49.