Question

Find the equation of the locus of a point the sum of whose distances from (0,2)and (0,-2) is 6
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Answer 1

GIVEN :–

• Sum of distances of a point from (0,2) and (0,-2) is 6.

TO FIND :–

• Locus of the point = ?

SOLUTION :–

• Let the point be (x , y) .

▪︎ Distance of the point from (0,2) is –

$\\ \implies { \bold{ d_{1} = \sqrt{ {(0 - x)}^{2} + {(2 - y)}^{2} } }}$

$\\ \implies { \bold{ d_{1} = \sqrt{ {(x)}^{2} + {(2 - y)}^{2} } \: \: \: \: \: \: - - - eq.(1)}}$

▪︎ Distance of the point from (0,-2) –

$\\ \implies { \bold{ d_{2} = \sqrt{ {(0 - x)}^{2} + {( - 2 - y)}^{2} } }}$

$\\ \implies { \bold{ d_{2} = \sqrt{ {(x)}^{2} + {(2 + y)}^{2} } \: \: \: \: \: \: - - - eq.(2)}}$

● According to the question –

$\\ \implies { \bold{ Sum \: \: of \: \: distances = 6}}$

$\\ \implies { \bold{ d_{1} + d_{2}= 6}}$

• Using eq.(1) and eq.(2) –

$\\ \implies { \bold{ \sqrt{ {(x)}^{2} + {(2 - y)}^{2} } + \sqrt{ {(x)}^{2} + {(2 + y)}^{2} } = 6}}$

• Square on both sides –

$\\ \implies { \bold{ \left(\sqrt{ {(x)}^{2} + {(2 - y)}^{2} } + \sqrt{ {(x)}^{2} + {(2 + y)}^{2} } \right)^{2} = {6}^{2} }}$

$\\ \implies { \bold{ { {(x)}^{2} + {(2 - y)}^{2} } +{ {(x)}^{2} + {(2 + y)}^{2} } + 2\sqrt{ {(x)}^{2} + {(2 - y)}^{2} }\sqrt{ {(x)}^{2} + {(2 + y)}^{2} }= 36 }}$

$\\ \implies { \bold{ { 2{x}^{2}} + 4 + {y}^{2} - 4y + 4 + {y}^{2} + 4y + 2\sqrt{ {(x)}^{2} + {(2 - y)}^{2} }\sqrt{ {(x)}^{2} + {(2 + y)}^{2} }= 36 }}$

$\\ \implies { \bold{ { 2{x}^{2}} + 8 + 2{y}^{2} + 2\sqrt{ {(x)}^{2} + {(2 - y)}^{2} }\sqrt{ {(x)}^{2} + {(2 + y)}^{2} }= 36 }}$

$\\ \implies { \bold{ { 2{x}^{2}} + 2{y}^{2} + 2\sqrt{ {(x)}^{2} + {(2 - y)}^{2} }\sqrt{ {(x)}^{2} + {(2 + y)}^{2} }= 28 }}$

$\\ \implies { \bold{ { {x}^{2}} + {y}^{2} + \sqrt{ {(x)}^{2} + {(2 - y)}^{2} }\sqrt{ {(x)}^{2} + {(2 + y)}^{2} }= 14 }}$

$\\ \implies { \bold{ { {x}^{2}} + {y}^{2} + \sqrt{ \{{(x)}^{2} + {(2 - y)}^{2} \} \{{(x)}^{2} + {(2 + y)}^{2}\}}= 14 }}$

$\\ \implies { \bold{ { {x}^{2}} + {y}^{2} + \sqrt{ {x}^{4} + {x}^{2} {(2 + y)}^{2} + {x}^{2} {(2 - y)}^{2} + { \{(2 - y)(2 + y) \}}^{2} }= 14 }}$

$\\ \implies { \bold{ { {x}^{2}} + {y}^{2} + \sqrt{ {x}^{4} + 2 {x}^{2} {(4 + {y}^{2} )} + {(4 - {y}^{2}) }^{2} }= 14 }}$

$\\ \implies { \bold{ { {x}^{2}} + {y}^{2} - 14 = \sqrt{ {x}^{4} + 2 {x}^{2} {(4 + {y}^{2} )} + {(4 - {y}^{2}) }^{2} }}}$

• Square on both sides –

$\\ \implies { \bold{ ( { {x}^{2}} + {y}^{2} - 14 )^{2} = { {x}^{4} + 2 {x}^{2} {(4 + {y}^{2} )} + {(4 - {y}^{2}) }^{2} }}}$

Answer 2

Given :-

•The sum of distances of a point from (0,2) , (0,-2) is 6 .

To Find :-

• The locus of a point.

Solution :-

• Consider, the two points are R and A.

•Distance of the point from (0,2) is,

=> D1 = √( 0 - R)² + (2 - A)²

=> D1 = √(R)² + (2 - A)²

•Distance of the point from (0,-2) is,

=> D2 = √(0 - R)² + (-2 - A)²

=> D2 = √(R)² + (2 + A)²

•Given that, Sum of distances = 6

=> .°. D1 + D2 = 6

[ Putting the values ]

=> √[(R)² + (2 - A)² + √(R)² + (2 + A)²] = 6

[ Square on each sides. ]

=> (√[(R)² + (2 - A)² + √(R)² + (2 + A)²])²=(6)²

=> (R)² + (2 - A)² + (R)² + (2 + A)² + 2√(R)² + (2 - A)² √(R)² + (2 + A)² = 36 .

=> 2R² + 4 + A² - 4A + 4 + A² + 4A + 2 + 2√(R)² + (2 - A)² √(R)² + (2 + A)² = 36

=> 2R²+8 +2A²+ 2√(R)² + (2 - A)²√(R)²+(2 + A)² = 36

=> 2R² + 2A² + 2√(R)² + (2 - A)²√(R)²+(2 + A)² = 28

=> R² + A² +√(R)² + (2 - A)²√(R)²+(2 + A)² = 14

=> R² + A² +√{(R)² + (2 - A)² } { (√R)² + (2 + A)²} = 14

• After solving we get,

=> R² + A² - 14 = √R^4 + 2R²(4 + A)² + (4 - A²)²

[ Square on each sides. ]

=>(R²+A²-14)²= R^4+2R²(4+A²)+(4 - A²)²

Hence, the locus of point is (R²+A²-14)² = R^4+2R²(4+A²)+(4 - A²)² .