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

True

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False

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Solution

The correct option is A True
Let two given points are $A (a e, 0)$ and $B (- a e, 0)$

Let coordinates of required point are $C (x, y)$

By distance formula,
$C A = \sqrt{{(x - a e)}^{2} + {(y - 0)}^{2}}$
$∴ C A = \sqrt{{(x - a e)}^{2} + y^{2}}$

Similarly, $C B = \sqrt{{(x - (- a e))}^{2} + {(y - 0)}^{2}}$
$∴ C B = \sqrt{{(x + a e)}^{2} + y^{2}}$

By given condition,
$\sqrt{{(x - a e)}^{2} + y^{2}} + \sqrt{{(x + a e)}^{2} + y^{2}} = 2 a$
$∴ \sqrt{{(x - a e)}^{2} + y^{2}} = 2 a - \sqrt{{(x + a e)}^{2} + y^{2}}$

Squaring both sides, we get,
${(\sqrt{{(x - a e)}^{2} + y^{2}})}^{2} = {(2 a - \sqrt{{(x + a e)}^{2} + y^{2}})}^{2}$
$∴ {(x - a e)}^{2} + y^{2} = {(2 a - \sqrt{{(x + a e)}^{2} + y^{2}})}^{2}$
$∴ x^{2} - 2 a e x + a^{2} e^{2} + y^{2} = 4 a^{2} - 4 a \sqrt{{(x + a e)}^{2} + y^{2}} + {(x + a e)}^{2} + y^{2}$
$∴ x^{2} - 2 a e x + a^{2} e^{2} + y^{2} = 4 a^{2} - 4 a \sqrt{{(x + a e)}^{2} + y^{2}} + x^{2} + 2 a e x + a^{2} e^{2} + y^{2}$
$∴ - 2 a e x = 4 a^{2} - 4 a \sqrt{{(x + a e)}^{2} + y^{2}} + 2 a e x$
$∴ 4 a \sqrt{{(x + a e)}^{2} + y^{2}} = 4 a^{2} + 2 a e x + 2 a e x$
$∴ 4 a \sqrt{{(x + a e)}^{2} + y^{2}} = 4 a^{2} + 4 a e x$
$∴ 4 a \sqrt{{(x + a e)}^{2} + y^{2}} = 4 a (a + e x)$
$∴ \sqrt{{(x + a e)}^{2} + y^{2}} = (a + e x)$

Squaring both sides, we get,
${(x + a e)}^{2} + y^{2} = {(a + e x)}^{2}$
$∴ x^{2} + 2 a e x + a^{2} e^{2} + y^{2} = a^{2} + 2 a e x + e^{2} x^{2}$
$∴ x^{2} - e^{2} x^{2} + y^{2} = a^{2} - a^{2} e^{2}$
$∴ x^{2} (1 - e^{2}) + y^{2} = a^{2} (1 - e^{2})$

Divide both sides by $a^{2} (1 - e^{2})$ , we get,
$∴ \frac{x^{2} (1 - e^{2})}{a^{2} (1 - e^{2})} + \frac{y^{2}}{a^{2} (1 - e^{2})} = \frac{a^{2} (1 - e^{2})}{a^{2} (1 - e^{2})}$
$∴ \frac{x^{2}}{a^{2}} + \frac{y^{2}}{a^{2} (1 - e^{2})} = 1$

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

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Similar questions

Q. 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})

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

Q. 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

Q. A point moves so that the difference of its distances from (ae, 0) and (−ae, 0) is 2a. Prove that the equation to its locus is

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

, where b² = a² (e² − 1).

A (0, 1)

and

B (0, - 1)

are

2

points. If a variable point P moves such that sum of its distance fro A and B is

4

. Then the locus of P is the equation of the form of

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

. Find the value of

(a^{2} + b^{2})

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Determine whether the equation of the locus of a point, which moves such that the sum of its distances from two given points (ae,0) and (−ae,0) is equal to 2a, is x2a2+y2b2=1

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