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Question: If P is any point lying on the ellipse \(\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1\), a \> b wit...

If P is any point lying on the ellipse x2a2+y2b2=1\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, a > b with foci S and S', then locus of the incentre of the triangle PSS' is

A

(1−e)x2+(1+e)y2=a2(1 - e)x^{2} + (1 + e)y^{2} = a^{2}

B

(1+e)x2+(1−e)y2=a2e2(1 + e)x^{2} + (1 - e)y^{2} = a^{2}e^{2}

C

(1+e)x2+(1−e)y2=a2(1 + e)x^{2} + (1 - e)y^{2} = a^{2}

D

(1−e)x2+(1+e)y2=a2e2(1 - e)x^{2} + (1 + e)y^{2} = a^{2}e^{2}

Answer

(1−e)x2+(1+e)y2=a2e2(1 - e)x^{2} + (1 + e)y^{2} = a^{2}e^{2}

Explanation

Solution

Let coordinates of P be (a cosθ, b sin θ),

SS' = 2ae, PS = a (1-e cosθ) and PS' = a(1+e cosθ)

∴ Coordinates of the incentre is given by

x=2ae.acos⁡θ−aea(1−ecos⁡θ)+ae.a(1+ecos⁡θ)2a(1+e)\frac{2ae.a\cos\theta - aea\left( 1 - e\cos\theta \right) + ae.a\left( 1 + e\cos\theta \right)}{2a(1 + e)}= a2e[2cos⁡θ−1+ecos⁡θ+1+ecos⁡θ]2a(1+e)=aecos⁡θ\frac{a^{2}e\left\lbrack 2\cos\theta - 1 + e\cos\theta + 1 + e\cos\theta \right\rbrack}{2a(1 + e)} = ae\cos\thetaand y =

2ae.bsin⁡θ2a(1+e)=be1+esin⁡θ\frac{2ae.b\sin\theta}{2a(1 + e)} = \frac{be}{1 + e}\sin\theta∴ Locus of the incentre of the

triangle PS'S is

x2a2e2+y2(1+e)2b2e2\frac{x^{2}}{a^{2}e^{2}} + \frac{y^{2}(1 + e)^{2}}{b^{2}e^{2}}=1

or x2a2e2+y2(1+e)2a2e2(1+e)(1−e)\frac{x^{2}}{a^{2}e^{2}} + \frac{y^{2}(1 + e)^{2}}{a^{2}e^{2}(1 + e)(1 - e)}=1

or (1−e)x2+(1+e)y2=a2e2(1 - e)x^{2} + (1 + e)y^{2} = a^{2}e^{2}