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  • If the curve  <img alt="\mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cfrac%7Bx%5E2%7D%7Ba%5E2%7D&plus;%5Cfrac%7By%5E2%7D%7

If the curve  \mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1} and \mathrm{\frac{x^2}{\alpha^2}-\frac{y^2}{\beta^2}=1} cut each other orthogonally then

Option: 1

\mathrm{a^2+b^2=\alpha^2+\beta^2}


Option: 2

\mathrm{a^2-b^2=\alpha^2-\beta^2}


Option: 3

\mathrm{a^2-b^2-\alpha^2+\beta^2}


Option: 4

\mathrm{a^2+b^2-\alpha^2-\beta^2}


Answers (1)

best_answer

\mathrm{\text { Slope of } \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \text { at } P\left(x_0, y_0\right) \text { is }-\frac{b^2 x_0}{a^2 y_0} \text {, Slope of } \frac{x^2}{\alpha^2}-\frac{y^2}{\beta^2}=1 \text { at } P\left(x_0, y_0\right) \text { is } \frac{\beta^2 x_0}{\alpha^2 y_0}}

\mathrm{M_1 M_2=-1 \Rightarrow b^2 \beta^2 x_0^2=a^2 \alpha^2 y_0^2}   ----------------(1)

Now solving the curves

\mathrm{x_6^2\left(\frac{1}{a^2}-\frac{1}{\alpha^2}\right)=-y_0^2\left(\frac{1}{b^2}+\frac{1}{\beta^2}\right)} -------------(2)

From (1)  and (2)

 \mathrm{\frac{a^2-\frac{1}{\alpha^2}}{\frac{1}{b^2}+\frac{1}{\beta^2}}=\frac{b^2 \beta^2}{a^2 \alpha^2} \Rightarrow a^2-b^2=\alpha^2+\beta^2}

Posted by

Kshitij

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