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  • The locus of the mid-points of the portion of the tangents to the ellipse <img alt="\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1" src="https://learn.careers360.com/latex-image/?%5Cfrac%7Bx%5E%7B2%7D%7

The locus of the mid-points of the portion of the tangents to the ellipse \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1intercepted between the axes is?

Option: 1

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


Option: 2

\frac{a^{2}}{x^{2}}+\frac{b^{2}}{y^{2}}=4


Option: 3

4 r^{2}=a^{2} \sin ^{2}\theta +b^{2} \cos ^{2} \theta


Option: 4

Both A and B


Answers (1)

best_answer

 

 

Pair of Tangent -

Pair of Tangent:

Where points Q and R are the point of contacts of the tangents to the ellipse.

-

 

 

The equation of any tangent to the ellipse at P(\theta) is

A B: \frac{x}{a} \cos \theta+\frac{y}{b} \sin \theta=1

\begin{array}{l}{\text { Thus } A=(a \sec \theta, 0) \text { and } A=(0, b \csc \theta)} \\ {\text { Let }(h, k) \text { be the mid-point of the tangent } A B .} \\ \end{array}

\text { Therefore, } h=\frac{a \sec \theta}{2} \text { and } k=\frac{b \csc \theta}{2}

\Rightarrow \quad \cos \theta=\frac{a}{2 h} \text { and } \sin \theta=\frac{b}{2 k}

Now squaring and adding, we get

       \frac{a^{2}}{4 h^{2}}+\frac{b^{2}}{4 k^{2}}=1

\Rightarrow \quad \frac{a^{2}}{h^{2}}+\frac{b^{2}}{k^{2}}=4

Hence the locus of (h, k) is

\frac{a^{2}}{x^{2}}+\frac{b^{2}}{y^{2}}=4

Now, put x=r \cos \theta \;\;\&\;\; y=r \sin \theta

we get, 4 r^{2}=a^{2} \sin ^{2}\theta +b^{2} \cos ^{2} \theta.

Posted by

sudhir.kumar

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