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

✰QuesTion :-

$\rm Find \: the \: equation \: of \: the \: locus \: of \: a \: point \: \\ </p><p> \rm the \: sum \: of \: whose \: distances \: from \: (0, 2) \: \\ \rm</p><p>and\: (0, -2) \: is \: 6 \: units.$

✰Answer :-

$\boxed{ \to \rm \: \frac{ {h}^{2} }{5} + \frac{ {k}^{2} }{9} = 1 }\\ \\ \sf this \: is \: \: equation \: of \: an \: ellipse \:$

✰To Find :-

→ Equation of locus.

✰Solution :-

According to the question,

→ distance of the sum of distance A(0,2) and B(0,-2) from locus is 6 units .

Let P( h, k) point on locus .

According to the condition -

→ PA + PB = 6 units ...... eq.(1)

Firstly find the distance between P and A ,by distance formula -

$\boxed{ \rm \: D = \sqrt{ {( x_2-x_1)}^{2} + {( y_2-y_1)}^{2} } } \\ \\ \sf \: hence \: \\ \rm distance \: between \: P( h, k) \: and \: A(0,2) \: is - \\ \\ \rm PA = \sqrt{ {(0 - h)}^{2} + {(2 - k)}^{2} } \\ \\ \mapsto \rm PA = \sqrt{ {h}^{2} + {4} + {k}^{2} - 4k } \: \: eq....(2) \\ \\ \rm now \: calculate \: distance \: between \: P( h, k) \: and \: \\ \rm B(0,-2) \: \: \\ \\ \mapsto \rm PB \: = \sqrt{ {(0 - h)}^{2} + {( - 2 - k)}^{2} } \\ \\ \mapsto \rm PB = \sqrt{ {h}^{2} + 4 + { {k}^{2} } + 4k } \: \: \: eq...(3) \\ \\ \rm now \: put \: the \: value \:in \: eq.(1) \: from \: eq.(1) \: and \\ \rm\: eq.(2) \\ \\ \: \: \: \: \: \: \: \: \: \: \: \boxed{ \because \rm \:PA + PB = 6 \: } \\ \therefore \\ \: \\ \to\{ \rm \sqrt{ {h}^{2} + {4} + {k}^{2} - 4k } + \sqrt{ {h}^{2} + 4 + { {k}^{2} } + 4k } \} \\ \: \: \: \: \: = 6 \: \\ \\ \to \rm \sqrt{ {h}^{2} + {4} + {k}^{2} - 4k } - 6 = \sqrt{ {h}^{2} + 4 + { {k}^{2} } + 4k } \\ \\ \bf \red{ squaring \: on \: both \: sides} \\ \\ \to \rm {h}^{2} + 4 + {k}^{2} - 4k + 36 - 12\sqrt{ {h}^{2} + {4} + {k}^{2} - 4k } \: \\ \: \: \rm \: \: \: \: = {h}^{2} + 4 + {k}^{2} + 4k \\ \\ \rm \to 36 - 8k = 12\sqrt{ {h}^{2} + {4} + {k}^{2} - 4k } \: \\ \\ \to \rm 3 - \frac{2}{3} k = \sqrt{ {h}^{2} + {4} + {k}^{2} - 4k } \: \\ \\ \bf \green{again \: squaring \: on \: both \: sides} \\ \\ \to \rm { \bigg(3 - \frac{2}{3} k \bigg)}^{2} = {h}^{2} + 4 + {k}^{2} - 4k \\ \\ \to \rm 9 + \frac{4}{9} {k}^{2} - 2 \times 3 \times \frac{2}{3} k = {h}^{2} + 4 + {k}^{2} - 4k \\ \\ \to \rm 9 - 4k = {h}^{2} + 4 + {k}^{2} - 4k - \frac{4}{9} {k}^{2} \\ \:$

$\to \rm {h}^{2} + {k}^{2} \bigg(1 - \frac{4}{9} \bigg) = 9 - 4 \\ \\ \to \rm \: {h}^{2} + \frac{5}{9} {k}^{2} = 5 \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \to \rm \: \frac{ {h}^{2} }{5} + \frac{ {k}^{2} }{9} = 1 }\\ \\ \sf this \: is \: \: equation \: of \: an \: ellipse \:$