Show that the curves

Question:

Show that the curves $\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}+\lambda_{1}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}+\lambda_{1}}=1$ and $\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}+\lambda_{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}+\lambda_{2}}=1$ interest at right angles

Solution:

Given:

Curves $\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}+\lambda_{1}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}+\lambda_{1}}=1 \ldots$ (1)

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

First curve is $\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}+\lambda_{1}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}+\lambda_{1}}=1$

Differentiating above w.r.t $x$,

$\Rightarrow \frac{2 x}{a^{2}+\lambda_{1}}+\frac{2 y}{b^{2}+\lambda_{1}} \cdot \frac{d y}{d x}=0$

$\Rightarrow \frac{y}{b^{2}+\lambda_{1}} \cdot \frac{d y}{d x}=\frac{-x}{a^{2}+\lambda_{1}}$

$\Rightarrow \frac{d y}{d x}=\frac{\frac{-x}{a^{2}+\lambda_{1}}}{\frac{y}{b^{2}+\lambda_{1}}}$

$\Rightarrow m_{1}=\frac{-x\left(b^{2}+\lambda_{1}\right)}{y\left(a^{2}+\lambda_{1}\right)} \ldots(3)$

Second curve is $\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}+\lambda_{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}+\lambda_{2}}=1$

Differentiating above w.r.t $x$,

$\Rightarrow \frac{2 x}{a^{2}+\lambda_{2}}+\frac{2 y}{b^{2}+\lambda_{2}} \cdot \frac{d y}{d x}=0$

$\Rightarrow \frac{y}{b^{2}+\lambda_{2}} \cdot \frac{d y}{d x}=\frac{-x}{a^{2}+\lambda_{2}}$

$\Rightarrow \frac{d y}{d x}=\frac{\frac{-x}{a^{2}+\lambda_{2}}}{\frac{y}{b^{2}+\lambda_{2}}}$

$\Rightarrow m_{2}=\frac{-x\left(b^{2}+\lambda_{2}\right)}{y\left(a^{2}+\lambda_{2}\right)} \ldots(4)$

Now equation (1) - (2) gives

$\Rightarrow\left(\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}+\lambda_{1}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}+\lambda_{1}}=1\right)-\left(\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}+\lambda_{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}+\lambda_{2}}=1\right)$

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

$\Rightarrow x^{2}\left(\frac{1}{a^{2}+\lambda_{1}}-\frac{1}{a^{2}+\lambda_{2}}\right)=-y^{2}\left(\frac{1}{b^{2}+\lambda_{1}}-\frac{1}{b^{2}+\lambda_{2}}\right)$

$\Rightarrow \frac{\mathrm{x}^{2}}{\mathrm{y}^{2}}=\frac{-\left(\frac{1}{\mathrm{~b}^{2}+\lambda_{1}}-\frac{1}{\mathrm{~b}^{2}+\lambda_{2}}\right)}{\left(\frac{1}{\mathrm{a}^{2}+\lambda_{1}}-\frac{1}{\mathrm{a}^{2}+\lambda_{2}}\right)}$

$\Rightarrow \frac{\mathrm{x}^{2}}{\mathrm{y}^{2}}=\frac{-\left(\frac{\mathrm{b}^{2}+\lambda_{2}-\left(\mathrm{b}^{2}+\lambda_{1}\right)}{\left(\mathrm{b}^{2}+\lambda_{1}\right)\left(\mathrm{b}^{2}+\lambda_{1}\right)}\right)}{\left(\frac{\left.\mathrm{a}^{2}+\lambda_{2}\right)-\mathrm{a}^{2}+\lambda_{1}}{\left(\mathrm{a}^{2}+\lambda_{1}\right)\left(\mathrm{a}^{2}+\lambda_{2}\right)}\right)}$

$\Rightarrow \frac{\mathrm{x}^{2}}{\mathrm{y}^{2}}=\frac{-\left(\frac{\mathrm{b}^{2}+\lambda_{2}-\mathrm{b}^{2}-\lambda_{1}}{\left(\mathrm{~b}^{2}+\lambda_{1}\right)\left(\mathrm{b}^{2}+\lambda_{1}\right)}\right)}{\left(\frac{\left(\mathrm{a}^{2}+\lambda_{2}-\mathrm{a}^{2}-\lambda_{1}\right.}{\left(\mathrm{a}^{2}+\lambda_{1}\right)\left(\mathrm{a}^{2}+\lambda_{2}\right)}\right)}$

$\Rightarrow \frac{\mathrm{x}^{2}}{\mathrm{y}^{2}}=\frac{\left(\frac{-\left(\lambda_{2}-\lambda_{1}\right)}{\left(\mathrm{b}^{2}+\lambda_{1}\right)\left(\mathrm{b}^{2}+\lambda_{1}\right)}\right)}{\left(\frac{\left(\lambda_{2}-\lambda_{1}\right.}{\left(\mathrm{a}^{2}+\lambda_{1}\right)\left(\mathrm{a}^{2}+\lambda_{2}\right)}\right)}$

$\Rightarrow \frac{\mathrm{x}^{2}}{\mathrm{y}^{2}}=\frac{\left(\frac{\left(\lambda_{1}-\lambda_{2}\right)}{\left(\mathrm{b}^{2}+\lambda_{1}\right)\left(\mathrm{b}^{2}+\lambda_{1}\right)}\right)}{\left(\frac{\left(\lambda_{2}-\lambda_{1}\right.}{\left(\mathrm{a}^{2}+\lambda_{1}\right)\left(\mathrm{a}^{2}+\lambda_{2}\right)}\right)}$

$\Rightarrow \frac{x^{2}}{y^{2}}=\frac{\left(\lambda_{1}-\lambda_{2}\right)\left(a^{2}+\lambda_{1}\right)\left(a^{2}+\lambda_{2}\right)}{\left(\lambda_{2}-\lambda_{1}\right)\left(b^{2}+\lambda_{1}\right)\left(b^{2}+\lambda_{1}\right)}$

$\Rightarrow \frac{x^{2}}{y^{2}}=\frac{-\left(\lambda_{2}-\lambda_{1}\right)\left(a^{2}+\lambda_{1}\right)\left(a^{2}+\lambda_{2}\right)}{\left(\lambda_{2}-\lambda_{1}\right)\left(b^{2}+\lambda_{1}\right)\left(b^{2}+\lambda_{1}\right)}$

$\Rightarrow \frac{x^{2}}{y^{2}}=\frac{-\left(a^{2}+\lambda_{1}\right)\left(a^{2}+\lambda_{2}\right)}{\left(b^{2}+\lambda_{1}\right)\left(b^{2}+\lambda_{1}\right)} \ldots(5)$

When $\mathrm{m}_{1}=\frac{-\mathrm{x}\left(\mathrm{b}^{2}+\lambda_{1}\right)}{\mathrm{y}\left(\mathrm{a}^{2}+\lambda_{1}\right)} \& \mathrm{~m}_{2}=\frac{-\mathrm{x}\left(\mathrm{b}^{2}+\lambda_{2}\right)}{\mathrm{y}\left(\mathrm{a}^{2}+\lambda_{2}\right)}$

$\Rightarrow \frac{-\mathrm{x}\left(\mathrm{b}^{2}+\lambda_{1}\right)}{\mathrm{y}\left(\mathrm{a}^{2}+\lambda_{1}\right)} \times \frac{-\mathrm{x}\left(\mathrm{b}^{2}+\lambda_{2}\right)}{\mathrm{y}\left(\mathrm{a}^{2}+\lambda_{2}\right)}$

$\Rightarrow \frac{\mathrm{x}^{2}}{\mathrm{y}^{2}} \times \frac{\left(\mathrm{b}^{2}+\lambda_{1}\right)\left(\mathrm{b}^{2}+\lambda_{2}\right)}{\left(\mathrm{a}^{2}+\lambda_{1}\right)\left(\mathrm{a}^{2}+\lambda_{2}\right)}$

Substituting $\frac{x^{2}}{y^{2}}$ from equation $(5)$, we get

$\Rightarrow \frac{-\left(\mathrm{a}^{2}+\lambda_{1}\right)\left(\mathrm{a}^{2}+\lambda_{2}\right)}{\left(\mathrm{b}^{2}+\lambda_{1}\right)\left(\mathrm{b}^{2}+\lambda_{1}\right)} \times \frac{\left(\mathrm{b}^{2}+\lambda_{1}\right)\left(\mathrm{b}^{2}+\lambda_{2}\right)}{\left(\mathrm{a}^{2}+\lambda_{1}\right)\left(\mathrm{a}^{2}+\lambda_{2}\right)}$

$\Rightarrow-1$

$\therefore$ The two curves intersect orthogonally,