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Section: Mathematics

Question : 15 of 27

Marks: +1, -0

x^2 + y^2 = 1

Solution:

Tagent at P:

x\cos\theta + y\sin\theta = 1

.....(i)

Tangent at S :

x = 1

....(ii)

∴ By (i) and (ii):Q

\left(1, \frac{1-\cos\theta}{\sin\theta}\right)

Line through Q parallel to RS:

y = \frac{1-\cos\theta}{\sin\theta} \Rightarrow y = \frac{\tan\theta}{2}

.......(iii)

Normal at P :

y = \frac{\sin\theta}{\cos\theta}x \Rightarrow y = \tan\theta \cdot x

......(iv)

Point of intersection of equation (iii) and (iv), E :

h = \frac{1 - \frac{\tan^2\theta}{2}}{2}; k = \frac{\tan\theta}{2}

eliminating θ:h

= \frac{1 - k^2}{2} \Rightarrow y^2 = 1 - 2x

Options (A) and (C) satisfies the locus.

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