Find the locus of a point which moves so that the difference of its distances from the points, $(5, 0)$ and $(- 5, 0)$ is $2$ is:

\frac{x^{2}}{1} + \frac{y^{2}}{24} = 1

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\frac{x^{2}}{24} + \frac{y^{2}}{1} = 1

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\frac{x^{2}}{24} - \frac{y^{2}}{2} = 1

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\frac{x^{2}}{1} - \frac{y^{2}}{24} = 1

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Solution

The correct option is B $\frac{x^{2}}{1} - \frac{y^{2}}{24} = 1$
The locus is nothing but hyperbola.
Difference of distance of a point from foci $= 2 a$
Given distance is $2 \Rightarrow a = 1$
Distance between foci $= 2 a e = 2 \sqrt{a^{2} + b^{2}} = \sqrt{(5 + 5)^{2}}$
$\Rightarrow a^{2} + b^{2} = 25$
$\Rightarrow b^{2} = 24$
Therefore, locus is $\frac{x^{2}}{1} - \frac{y^{2}}{24} = 1$

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Similar questions

The circle $x^{2} + y^{2} - 8 x = 0$ and hyperbola $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$ intersect at the points A and B.
Equation of the circle with AB as its diameter is

Q. The circle

x^{2} + y^{2} - 8 x = 0

and hyperbola

\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1

intersect at point

A

and

B

. The equation of circle with

A B

as diameter is:

Q. Find the shortest and the longest distances from the point

(1, 2, - 1)

to the sphere

x^{2} + y^{2} z^{2} = 24

List- I	List- II
A. The locus of the point of intersection of the perpendicular tangents of $x^{2} + y^{2} = a^{2}$	1) $x^{2} + y^{2} = 5$ of
B. The locus of the mid-point of the chord of the circle $x^{2} + y^{2} = 8$ which is $\sqrt{2}$ units from origin is	2) $x^{2} + y^{2} = a^{2} {sec}^{2} α$
C. The locus of the point of intersection of the lines $x = \frac{1 - t^{2}}{1 + t^{2}}$ and $y = \frac{2 t}{1 + t^{2}}$	3) $x^{2} + y^{2} = 2 a^{2}$
D. The locus of the point whose chord of contact w.r.t $x^{2} + y^{2} = a^{2}$ making an angle '2 $α$ ' at the center is	4) $x^{2} + y^{2} = 2$ 5) $x^{2} + y^{2} = 1$

Q. Show that, if

x^{2} + y^{2} = 1

, then the point

(\sqrt{x^{2} + y^{2}}, \sqrt{1 - x^{2} - y^{2}})

is at a distance

1

unit from the origin.

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Find the locus of a point which moves so that the difference of its distances from the points, (5,0) and (−5,0) is 2 is:

The correct option is B x21−y224=1The locus is nothing but hyperbola.Difference of distance of a point from foci =2a Given distance is 2⇒a=1Distance between foci =2ae=2√a2+b2=√(5+5)2 ⇒a2+b2=25 ⇒b2=24Therefore, locus is x21−y224=1

Find the locus of a point which moves so that the difference of its distances from the points, $(5, 0)$ and $(- 5, 0)$ is $2$ is: