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The tangent at any point \mathrm{P} of a given circle meets the tangents at a fixed point \mathrm{\mathrm{A}} in T, and T is joined to B, the other end of the diameter through A, The locus of the intersection of \mathrm{\mathrm{AP} \, \, and \, \, \mathrm{BT} } is an ellipse whose eccentricity is \mathrm{1 / \sqrt{2}. }

Option: 1

\frac{1}{2}


Option: 2

\frac{1}{3}


Option: 3

\frac{1}{\sqrt{5}}


Option: 4

\frac{1}{\sqrt{2}}


Answers (1)

best_answer

Tangents at \mathrm{P(a \cos \theta, a \sin \theta) } to
\mathrm{\begin{aligned} & x^2+y^2=a^2 \\ & x \cos \theta+y \sin \theta=a......................(1) \end{aligned} }
Equation of AP,
\mathrm{y=\frac{\sin \theta}{\cos \theta-1}(x-a) }...............................(2)
from (1) coordinates of the point
\mathrm{\mathrm{T} \equiv\left(\mathrm{a}, \frac{\mathrm{a}(1-\cos \theta)}{\sin \theta}\right) }

Equation of BT,
\mathrm{y=\frac{1-\cos \theta}{2 \sin \theta}(x+a) }...................................(3)
(h, k) be the point of intersection of equation
(2) and (3), then we get
\mathrm{\begin{aligned} & \mathrm{k}^2=-\frac{1}{2} \quad\left(\mathrm{~h}^2-\mathrm{a}^2\right) \\ \Rightarrow \quad & \frac{\mathrm{h}^2}{\mathrm{a}^2}+\frac{\mathrm{k}^2}{\mathrm{a}^2 / 2}=1 \end{aligned} }
Therefore locus of (h, k) is \mathrm{\frac{x^2}{a^2}+\frac{y^2}{a^2 / 2}=1 } Which is an ellipse for which \mathrm{e=\frac{1}{\sqrt{2}} }


 

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