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  • From a point on the circle , two tangents are drawn to the ell

From a point on the circle \mathrm{ x^2+y^2=1}, two tangents are drawn to the ellipse \mathrm{ a x^2+b y^2=1}. The locus of the midpoint of their chord of contact is \mathrm{\left(a x^2+b y^2\right)^2=k\left(x^2+y^2\right)}, where \mathrm{k=}

Option: 1

1


Option: 2

2


Option: 3

3


Option: 4

4


Answers (1)

best_answer

Let any point on the circle be\mathrm{ \mathrm{P}(\cos \theta, \sin \theta)} equation of chord of contact of tangents drawn from \mathrm{(\cos \theta, \sin \theta)} is givne by \mathrm{T=0}, so

ax \mathrm{\cos \theta+ by \sin \theta=1}       ......(1)

Again the equation of chord of an ellipse whose midpoint is \mathrm{(\alpha, \beta)} is given by \mathrm{T=S_1} so.

\mathrm{ a \alpha x+b \beta y-1=a \alpha^2+b \beta^2-1 }

\mathrm{ a \alpha x+b \beta y=a a^2+b \beta^2 }                ......(2)

Equation (1) and (2) represent the same line

therefore, \mathrm{\frac{a \cos \theta}{a \alpha}=\frac{b \sin \theta}{b \beta}=\frac{1}{a \alpha^2+b \beta^2}}

\mathrm{ \begin{gathered} \Rightarrow \cos \theta=\frac{\alpha}{\mathrm{a} \alpha^2+\mathrm{b} \beta^2} \text { and } \sin \theta=\frac{\beta}{\mathrm{a} \alpha^2+\mathrm{b} \beta^2} \\\\ \Rightarrow \frac{\alpha^2}{\left(\mathrm{a} \alpha^2+\mathrm{b} \beta^2\right)^2}+\frac{\beta^2}{\left(\mathrm{a} \alpha^2+\mathrm{b} \beta^2\right)^2}=1 \Rightarrow\left(\mathrm{a} \alpha^2+\mathrm{b} \beta^2\right)^2=\alpha^2+\beta^2 \end{gathered} }

\mathrm{\Rightarrow\left(a x^2+b y^2\right)^2=x^2+y^2}is the locus of the midpoint of the chord.

Posted by

Divya Prakash Singh

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