The two points A and B in a plane are such that for all points P lies on circle satisfied PA - PB - k- then k will not be equal toA- 3B- 2C- 0D- 1

The correct option is D 1 Let A = (a, 0), B( a, 0), P = (α, β) ∴PA^2/PB^2 = k^2 ⇒ (α a)^2 + β^2 = k^2[(α + a)^2 + β^2] ∴ Locus is (x^2 + y^2)(1 k^2) 2a(1 ...

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Standard XI

Mathematics

Terminologies Related to a Parabola

The two point...

Question

The two points A and B in a plane are such that for all points P lies on circle satisfied $\frac{P A}{P B} = k$ , then k will not be

equal to

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Solution

The correct option is B
1

Let A = (a, 0), B(-a, 0), P = $(α, β)$

$∴ \frac{P A^{2}}{P B^{2}} = k^{2} \Rightarrow (α - a)^{2} + β^{2} = k^{2} [(α + a)^{2} + β^{2}]$

$∴$ Locus is $(x^{2} + y^{2}) (1 - k^{2}) - 2 a (1 + k^{2}) x + (1 - k^{2}) a^{2} = 0$

This is a circle for k $\neq$ 1.

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Similar questions

The two points A and B in a plane are such that for all points P lies on circle satisfied $\frac{P A}{P B} = k$ , then k will not be equal to

Q. If A and B are fixed points in the plane such that

\frac{P A}{P B} = k

(constant) for all P on a given circle, then the value of k cannot be equal to

Q. Let

A

and

B

be two fixed points and

P

, another point in the plane, moves in such a way that

k_{1} P A + k_{2} P B = k_{3}

, where

k_{1}, k_{2}, k_{3}

are real constants. The locus of

P

Q. Assertion :If

A

and

B

are two fixed points and

P

a variable point, the locus of

P

such that

| P A - P B | = k

is a hyperbola for all

k

. Reason: If

S_{1}

and

S_{2}

are foci of a hyperbola and P be a point on the hyperbola then

| P S_{1} - P S_{2} | < S_{1} S_{2}

Q. If

A (1, 1)

and

B (2, - 3)

are two points and

P (h, k)

lies on the line

A B

such that

P A = 3 A B

, where

B

lies in between

A

and

P

, then the value of

h - k

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The two points A and B in a plane are such that for all points P lies on circle satisfied PAPB=k, then k will not be equal to

The correct option is B 1 Let A = (a, 0), B(-a, 0), P = (α,β) ∴PA2PB2=k2⇒(α−a)2+β2=k2[(α+a)2+β2] ∴ Locus is (x2+y2)(1−k2)−2a(1+k2)x+(1−k2)a2=0 This is a circle for k ≠ 1.

The two points A and B in a plane are such that for all points P lies on circle satisfied $\frac{P A}{P B} = k$ , then k will not be

equal to

The correct option is B
1

Let A = (a, 0), B(-a, 0), P = $(α, β)$

$∴ \frac{P A^{2}}{P B^{2}} = k^{2} \Rightarrow (α - a)^{2} + β^{2} = k^{2} [(α + a)^{2} + β^{2}]$

$∴$ Locus is $(x^{2} + y^{2}) (1 - k^{2}) - 2 a (1 + k^{2}) x + (1 - k^{2}) a^{2} = 0$

This is a circle for k $\neq$ 1.