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  • The locus of the point, the chord of contact of tangents from which 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

The locus of the point, the chord of contact of tangents from which to the ellipse \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 subtends a right angle at the centre of the ellipse is

Option: 1

\large \left(\frac{x^{2}}{a^{4}}+\frac{y^{2}}{b^{4}}\right)=\frac{1}{a}+\frac{1}{b}


Option: 2

\large \left(\frac{x^{2}}{a^{4}}+\frac{y^{2}}{b^{4}}\right)=\frac{1}{a^{4}}+\frac{1}{b^{4}}


Option: 3

\large \left(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\right)=\frac{1}{a^{2}}+\frac{1}{b^{2}}


Option: 4

\large \left(\frac{x^{2}}{a^{4}}+\frac{y^{2}}{b^{4}}\right)=\frac{1}{a^{2}}+\frac{1}{b^{2}}


Answers (1)

best_answer

Let the point from which tangents are drawn be (h, k). The equation of the chord of contact from the point (h, k) to the given ellipse is \frac{h x}{a^{2}}+\frac{k y}{b^{2}}=1

It subtends a right angle at the centre (0, 0) of the ellipse \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1

From the above two equation

\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=\left(\frac{x h}{a^{2}}+\frac{y k}{b^{2}}\right)^{2}

\Rightarrow\left(\frac{1}{a^{2}}-\frac{h^{2}}{a^{4}}\right) x^{2}+\left(\frac{1}{b^{2}}-\frac{k^{2}}{b^{4}}\right) y^{2}-\frac{2 h x k y}{a^{2} b^{2}}=0

Since these lines are at right angles, therefore sum of the co-efficients of x2 and y2 is zero.

\large \begin{array}{l}{\Rightarrow \quad\left(\frac{1}{a^{2}}-\frac{h^{2}}{a^{4}}\right)+\left(\frac{1}{b^{2}}-\frac{k^{2}}{b^{4}}\right)=0} \\\\ {\Rightarrow \quad\left(\frac{h^{2}}{a^{4}}+\frac{k^{2}}{b^{4}}\right)=\frac{1}{a^{2}}+\frac{1}{b^{2}}}\end{array}

Hence, the locus of (h, k) is

\left(\frac{x^{2}}{a^{4}}+\frac{y^{2}}{b^{4}}\right)=\frac{1}{a^{2}}+\frac{1}{b^{2}}

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