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  • The tangent at any point P of a given circle meets the tangent at a fixed point A in T, and T is joined to B, the other end of the diameter through A, The locus of the intersection o

The tangent at any point P of a given circle meets the tangent at a fixed point A in T, and T is joined to B, the other end of the diameter through A, The locus of the intersection of AP and BT is an ellipse whose eccentricity is.

Option: 1

\frac{1}{2}


Option: 2

\frac{1}{\sqrt{3}}


Option: 3

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


Option: 4

\frac{\sqrt{3}}{2}


Answers (1)

best_answer

Let equation of circle be \mathrm{x^2+y^2=a^2}    .....(1)

Any point P on (1) is \mathrm{(a \cos \theta, a \sin \theta)} Equation of tangent at P is

\mathrm{x \cos \theta+y \sin \theta=a}                     .....(2)

equation of tangent at A is \mathrm{x=a}        .....(3)

solving (1) and (2)

\mathrm{y=\frac{a(1-\cos \theta)}{\sin \theta}=\frac{2 a \sin ^2 \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}=a \tan \frac{\theta}{2} \Rightarrow T \equiv(a, a \tan \theta / 2)}
Equation of BT is 

\mathrm{y-0=\frac{a \tan \frac{\theta}{2}}{a+a}(x+a)}
\mathrm{ \Rightarrow \tan \frac{\theta}{2}=\frac{2 y}{x+a} }                                           .....(3)
Equation of AP is, \mathrm{y-0=\frac{a \sin \theta}{a \cos \theta-a}(x-a)}

\mathrm{ \begin{aligned} & \frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \sin ^2 \theta}(x-a) \\ & \Rightarrow y=-2 \sin ^2 \frac{\theta}{2} \quad \Rightarrow \cot \frac{\theta}{2}=-\frac{y}{x-a} \\ & \end{aligned} }                .....(4)
multiplying (3) and (4), we get

\mathrm{1=-\frac{2 y^2}{x^2-a^2} \Rightarrow x^2+2 y^2=a^2 \Rightarrow\frac{x^2}{a^2}+\frac{y^2}{\frac{a^2}{2}}=1}

which is equation of ellipse.

Now, 

\mathrm{\frac{a^2}{2}=a^2\left(1-e^2\right) \Rightarrow e=\frac{1}{\sqrt{2}}.}

Posted by

manish

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