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Question: At a point P on the ellipse \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1 tangent PQ is draw...

At a point P on the ellipse x2a2\frac{x^{2}}{a^{2}} + y2b2\frac{y^{2}}{b^{2}} = 1 tangent PQ is drawn. If the point Q be at a distance 1p\frac{1}{p} from the point P, where ‘p’ is distance of the tangent from the origin, then the locus of the point Q is –

A

x2a2+y2b2\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 +1a2b2\frac{1}{a^{2}b^{2}}

B

x2a2–y2b2\frac{x^{2}}{a^{2}}–\frac{y^{2}}{b^{2}} = 1 –1a2b2\frac{1}{a^{2}b^{2}}

C

x2a2+y2b2\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1a2b2\frac{1}{a^{2}b^{2}}

D

x2a2–y2b2\frac{x^{2}}{a^{2}}–\frac{y^{2}}{b^{2}} = 1a2b2\frac{1}{a^{2}b^{2}}

Answer

x2a2+y2b2\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 +1a2b2\frac{1}{a^{2}b^{2}}

Explanation

Solution

Equation of the tangent at P is

x−acos⁡θasin⁡θ\frac{x - a\cos\theta}{a\sin\theta} = y−bsin⁡θ−bcos⁡θ\frac{y - b\sin\theta}{- b\cos\theta}

The distance of the tangent from the origin is

P = ∣abb2cos⁡2θ+a2sin⁡2θ∣\left| \frac{ab}{\sqrt{b^{2}\cos^{2}\theta + a^{2}\sin^{2}\theta}} \right|

Ž 1p\frac{1}{p}= b2cos⁡2θ+a2sin⁡2θab\frac{\sqrt{b^{2}\cos^{2}\theta + a^{2}\sin^{2}\theta}}{ab}

Now the coordinates of the point Q are given as follows :

x−acos⁡θ−asin⁡θb2cos⁡2θ+a2sin⁡2θ\frac{\frac{x - a\cos\theta}{- a\sin\theta}}{\sqrt{b^{2}\cos^{2}\theta + a^{2}\sin^{2}\theta}}= y−bsin⁡θbcos⁡θb2cos⁡2θ+a2sin⁡2θ\frac{\frac{y - b\sin\theta}{b\cos\theta}}{\sqrt{b^{2}\cos^{2}\theta + a^{2}\sin^{2}\theta}}

= 1p\frac{1}{p} = b2cos⁡2θ+a2sin⁡2θab\frac{\sqrt{b^{2}\cos^{2}\theta + a^{2}\sin^{2}\theta}}{ab}

Ž x = a cos q – asin⁡θab\frac{a\sin\theta}{ab} and y = b sin q + bcos⁡θab\frac{b\cos\theta}{ab}.

Ž (xa)2+(yb)2\left( \frac{x}{a} \right)^{2} + \left( \frac{y}{b} \right)^{2}= 1 1a2b2\frac{1}{a^{2}b^{2}} is the required locus.

Hence (1) is the correct answer.