The equation of locus of $P$ , if $A = (4, 0), B (- 4, 0)$ and $| P A + P B | = 4$ is

3 x^{2} - y^{2} = 12

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x^{2} + y^{2} = 4

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3 x^{2} + y^{2} = 9

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x^{2} + 3 y^{2} = 20

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Solution

The correct option is A $3 x^{2} - y^{2} = 12$
Let $P \equiv (x, y)$
So,
$\sqrt{{(x - 4)}^{2} + y^{2}} + \sqrt{{(x + 4)}^{2} + y^{2}} = 4 {(x - 4)}^{2} + y^{2} = 16 + {(x + 4)}^{2} + y^{2} - 8 \sqrt{{(x + 4)}^{2} + y^{2}} 8 \sqrt{{(x + 4)}^{2} + y^{2}} = 16 + 16 x {(x + 4)}^{2} + y^{2} = 4 + 4 x^{2} + 8 x x^{2} + 16 + 8 x + y^{2} = 4 + 4 x^{2} + 8 x 3 x^{2} - y^{2} = 12 \frac{x^{2}}{4} - \frac{y^{2}}{12} = 1$

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The equation of locus of P, if A=(4, 0),B(−4, 0) and |PA+PB|=4 is

The correct option is A 3x2−y2=12Let P≡(x,y)So, √(x−4)2+y2+√(x+4)2+y2=4(x−4)2+y2=16+(x+4)2+y2−8√(x+4)2+y28√(x+4)2+y2=16+16x(x+4)2+y2=4+4x2+8xx2+16+8x+y2=4+4x2+8x3x2−y2=12x24−y212=1

The equation of locus of $P$ , if $A = (4, 0), B (- 4, 0)$ and $| P A + P B | = 4$ is