Let P=(1,0) and Q=(- 1,0) are two fixed points and R is a variable point on one side of the line PQ such that angle RPQ- angle RQP=45o, then the locus of R is

Question

Let P=(1,0) and Q=(- 1,0) are two fixed points and R is a variable point on one side of the line PQ such that angle RPQ- angle RQP=45o, then the locus of R is (A) y2-x2+2xy-1=0 (B) x2-y2+2xy+1=0 (C) y2+x2-2xy=1 (D) y2-x2-2xy+1=0

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/m-e3nvvnyonkyx83f6.jpg" /> Let, angle RPQ=θ & angle RQP=φ , where θ -φ=45° Now, tan(θ - φ)=1 ⇒ (tan θ - tan â¡ φ/1 + tan â¡ θ tan â¡ φ)=1 , where tan θ =(y/1 - x) &tanâ¡φ=(y/x + 1) ⇒ (y/1 - x)-(y/x + 1)=1+((y/1 - x))((y/x + 1)) ⇒ 2yx=1-x2+y2 ⇒ y2-x2-2xy+1=0

Q. Let $P = (1, 0)$ and $Q = (- 1, 0)$ are two fixed points and $R$ is a variable point on one side of the line $PQ$ such that $∠ RPQ - ∠ RQP = 4 5^{o},$ then the locus of $R$ is

$y^{2} - x^{2} + 2 x y - 1 = 0$

$x^{2} - y^{2} + 2 x y + 1 = 0$

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

$y^{2} - x^{2} - 2 x y + 1 = 0$

Q. Let P=(1,0) and Q=(−1,0) are two fixed points and R is a variable point on one side of the line PQ such that ∠RPQ−∠RQP=45o, then the locus of R is

y2−x2+2xy−1=0

x2−y2+2xy+1=0

y2+x2−2xy=1

y2−x2−2xy+1=0

Let, ∠RPQ=θ & ∠RQP=ϕ , where θ−ϕ=45∘ Now, tan(θ−ϕ)=1 ⇒1+tan⁡θtan⁡ϕtanθ−tan⁡ϕ​=1 , where tanθ=1−xy​&tan⁡ϕ=x+1y​ ⇒1−xy​−x+1y​=1+(1−xy​)(x+1y​) ⇒2yx=1−x2+y2 ⇒y2−x2−2xy+1=0

Q. Let $P = (1, 0)$ and $Q = (- 1, 0)$ are two fixed points and $R$ is a variable point on one side of the line $PQ$ such that $∠ RPQ - ∠ RQP = 4 5^{o},$ then the locus of $R$ is

$y^{2} - x^{2} + 2 x y - 1 = 0$

$x^{2} - y^{2} + 2 x y + 1 = 0$

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

$y^{2} - x^{2} - 2 x y + 1 = 0$