The equation of the locus of a point, which moves so that the sum of its distance from two given points $(a e, 0)$ and $(- a e, 0)$ is equal to $2 a$ , where $b^{2} = a^{2} (1 - e^{2})$ is

\frac{x^{2}}{a^{2}} + \frac{x^{2}}{b^{2}} = 1

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\frac{x^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1

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\frac{x^{2}}{a} + \frac{y^{2}}{b^{2}} = 1

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Solution

The correct option is C $\frac{x^{2}}{a} + \frac{y^{2}}{b^{2}} = 1$
$(a e, 0) (- a e, 0)$ distance $2 a$
$\sqrt{(x - a e)^{2} + y^{2}} + \sqrt{(x + a e)^{2} + y^{2}} = 2 a$
$\Rightarrow (x - a e)^{2} + y^{2} + (x + a e)^{2} + y^{2} + 2 \sqrt{(x^{2} + a^{2} e^{2} + 2 a e x) - y^{2})} (x^{2} - 2 a e x + a^{2} e^{2} + y^{2} = 4 a^{2}$
$\Rightarrow 2 x^{2} + 2 a^{2} e^{2} + 2 y^{2} + 2 \sqrt{(x^{2} + a^{2} e^{2} + y^{2})^{2} - (a^{2} + 4 a^{2} e^{2} x^{2})} = 4 a^{2}$
$(2 a^{2} - x^{2} - a^{2} e^{2} - y^{2})^{2} = (x^{2} + a^{2} e^{2} + y^{2})^{2} - 4 a^{2} e^{2} x^{2}$
$(2 a^{2} - x^{2} - a^{2} e^{2} - a^{2} e^{2} - y^{2})^{2} = x^{2} + a^{2} e^{2} + y^{2} - 2 a^{2} e^{2} x^{2}$
$4 a^{2} e^{2} x^{2} = (x^{2} + a^{2} e^{2} + y^{2} - x^{2} y^{2} - a^{2} e^{2} + 2 a^{2}) (x^{2} + a^{2} e^{2} x^{2} + x^{2} + y^{2} + a^{2} e^{2} - 2 a^{2})$
$4 a^{2} e^{2} x^{2} = 2 a^{2} (2 x^{2} + (a^{2} e^{2} + 1) y^{2} + a^{2} e^{2} - 2 a^{2}$
$2 e^{2} x^{2} = 2 x^{2} + (a^{2} e^{2} + 1) y^{2} + a^{2} e^{2} - 2 a^{2}$
by solving $\Rightarrow \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$

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Similar questions

Q. Determine whether the equation of the locus of a point, which moves such that the sum of its distances from two given points

(a e, 0) a n d (- a e, 0)

is equal to

2 a

, is

\frac{x^{2}}{a^{2}}

\frac{y^{2}}{b^{2}}

$A point moves as so that the difference of its distances from (ae,0)and(-ae,0)is 2 a, Prove that the equation to its locus is \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1, w h e r e b^{2} = a^{2} (e^{2} - 1) .$

Q. Find the area bounded by the ellipse

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

and the ordinates

x = 0

and

x = a e

, where

b^{2} = a^{2} (1 - e^{2})

and

e < 1

Q. A points moves that the sum of its distances from two fixed points (ae , 0) and (-ae , 0 ) is always 2a . Prove that the equation of the locus is

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{a^{2} (1 - e^{2})} = 1

p (θ), D (θ + \frac{π}{2})

are two points on the Ellipse

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

Then the locus of point of intersection of the two tangents at P and D to the ellipse is

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The equation of the locus of a point, which moves so that the sum of its distance from two given points (ae,0) and (−ae,0) is equal to 2a, where b2=a2(1−e2) is

The equation of the locus of a point, which moves so that the sum of its distance from two given points $(a e, 0)$ and $(- a e, 0)$ is equal to $2 a$ , where $b^{2} = a^{2} (1 - e^{2})$ is