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#### 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 thenOption: 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 $\mathrm{a^2+b^2-\alpha^2-\beta^2}$

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

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

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