Question

Given P= (a,0) and Q= (-a,0) and R is a variable point on one side of the line PQ such that ∠RPQ−∠RQP=2α. the locus of the point R is :-

Answer 1

Gɪᴠᴇɴ :-

• Coordinates of point P be (a, 0) and Q be (- a, 0).

• R be a variable point on one side of PQ such that ∠RPQ−∠RQP=2α.

Tᴏ Fɪɴᴅ :

• The locus of point R.

Construction :-

Through R, draw RS perpendicular to PQ intersecting PQ at S.

Cᴀʟᴄᴜʟᴀᴛɪᴏɴ :

Let assume that coordinates of R be (x, y).

So,

• OS = x

• RS = y

• QS = OS + OQ = a + x

• PS = OP - OS = a - x

Let

• ∠RPQ = d

• ∠RQP = c

Thus,

$\rm :\longmapsto\:2 \alpha = d - c$

$\rm :\implies\:tan2 \alpha = tan(d - c)$

$\bf:\longmapsto\:tan2 \alpha = \dfrac{tand - tanc}{1 + tanc \: tand} - - - (1)$

$\blue{\boxed{ \quad \because \: \bf \: tan(x - y) = \dfrac{tanx - tany}{1 + tanx \: tany} \quad \quad}}$

Now, we have to find the value of tanc and tand,

So,

Consider,

$\rm :\longmapsto\:In \: \triangle \: RQS$

$\rm :\longmapsto\:tanc = \dfrac{RS}{QS}$

$\bf :\longmapsto\: \purple{\boxed{ \bf \: tanc = \dfrac{y}{a + x}}} - - - (2)$

Consider,

$\rm :\longmapsto\:In \: \triangle \: RPS$

$\rm :\longmapsto\:tand = \dfrac{RS}{PS}$

$\bf :\longmapsto\: \purple{\boxed{ \bf \: tand = \dfrac{y}{a - x}}} - - - (3)$

Now,

On substituting the values of tanc and tand in equation (1),

$\rm :\longmapsto\:tan2 \alpha = \dfrac{\dfrac{y}{a - x} - \dfrac{y}{a + x} }{ \: \: \: \: 1 + \dfrac{y}{a - x} \times \dfrac{y}{a + x} \: \: \: \: }$

$\rm :\longmapsto\:tan2 \alpha = \dfrac{\dfrac{y(a + x) - y(a - x)}{ \cancel{(a - x)(a + x)}}}{ \: \: \: \: \dfrac{(a + x)(a - x) + {y}^{2} }{ \cancel{(a - x)(a + x)}} \: \: \: \: }$

$\rm :\longmapsto\:tan2 \alpha = \dfrac{y( \cancel{a} + x - \cancel{a} + x)}{ \: \: \: {a}^{2} - {x}^{2} + {y}^{2} \: \: \: }$

$\rm :\longmapsto\:tan2 \alpha = \dfrac{2xy}{ \: \: \: {a}^{2} - {x}^{2} + {y}^{2} \: \: \: }$

$\rm :\longmapsto\:({a}^{2} - {x}^{2} + {y}^{2})tan2 \alpha = 2xy$

$\rm :\longmapsto\:2xy - ({a}^{2} - {x}^{2} + {y}^{2})tan2 \alpha = 0$

$\bf :\longmapsto\:({x}^{2} - {y}^{2} - {a}^{2})tan2 \alpha + 2xy = 0$

Or

$\bf:\longmapsto\:{x}^{2}-{y}^{2}-{a}^{2} + 2xy\:cot2\alpha=0$

Hence,

$\underbrace{\purple{ \boxed{ \tt \: Locus\:of\: R \:is\rm\: \: {x}^{2}-{y}^{2}-{a}^{2} + 2xy\:cot2\alpha=0}}}$

Answer 2

Step-by-step explanation:

InΔRMP,

tanθ=

MP

RM

=

a−x

1

y

1

InΔRQM,

tanϕ=

QM

RM

=

a+x

1

y

1

\end{align}$$

And also given that,

∠RPQ−∠RQP=2α

θ−ϕ=2α

Taking tan both side and we get,

tan(θ−ϕ)=tan2α

⇒

1+tanθtanϕ

tanθ−tanϕ

=tan2α

⇒tan2α=

1+(

a−x

1

y

1

)(

a+x

1

y

1

)

(

a−x

1

y

1

)−(

a+x

1

y

1

)

⇒tan2α=

(a−x

1

)(a+x

1

)

(a−x

1

)(a+x

1

)+y

1

2

(a−x

1

)(a+x

1

)

ay

1

+x

1

y

1

−ay

1

+x

1

y

1

⇒tan2α=

a

2

−x

1

2

+y

1

2

2x

1

y

1

⇒

cot2α

1

=

a

2

−x

1

2

+y

1

2

2x

1

y

1

⇒a

2

−x

1

2

+y

1

2

=2x

1

y

1

cot2α

⇒a

2

−x

1

2

+y

1

2

−2x

1

y

1

cot2α=0

⇒x

1

2

−y

1

2

+2x

1

y

1

cot2α−a

2

=0

Hence, the locus of this equation is

x

2

−y

2

+2xycot2α−a

2

=0

x

2

−y

2

+2xycot2α=a

2