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  • If the chords of contact of tangents from two points <img alt="\mathrm{\left(x_{1}, y_{1}\right)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cleft%28x_%7B1%7D%2C%20y_%7B1%7

If the chords of contact of tangents from two points \mathrm{\left(x_{1}, y_{1}\right)} and \mathrm{\left(x_{2}, y_{2}\right)}  to the ellipse \mathrm{\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1}are at right angles then \mathrm{\left(\frac{\mathrm{x}_{1} \mathrm{x}_{2}}{\mathrm{y}_{1} \mathrm{y}_{2}}\right)} is equal to

Option: 1

\frac{a^{2}}{b^{2}}


Option: 2

-\frac{b^{2}}{a^{2}}


Option: 3

-\frac{a^{4}}{b^{4}}


Option: 4

-\frac{b^{4}}{a^{4}}


Answers (1)

best_answer

Chord of contact from \mathrm{\left(x_{1}, y_{1}\right)} is
\mathrm{\frac{x x_{1}}{a^{2}}+\frac{y y_{1}}{b^{2}}=1}

Whose slope is \mathrm{-\frac{b^{2}}{a^{2}} \frac{x_{1}}{y_{1}}}
Similarly slope of another chord of contact is \mathrm{-\frac{b^{2}}{a^{2}} \frac{x_{2}}{y_{2}}}
We have \mathrm{\left(-\frac{b^{2}}{a^{2}} \frac{x_{1}}{y_{1}}\right) \times\left(-\frac{b^{2}}{a^{2}} \frac{x_{2}}{y_{2}}\right)=-1 \Rightarrow \frac{x_{1} x_{2}}{y_{1} y_{2}}=-\frac{a^{4}}{b^{4}}}

Posted by

Deependra Verma

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