Question

Show that the following set of curves intersect orthogonally.
$x^2+4y^2=8$ and $x^2-2y^2=4$

Answer 1

$x^2+4y^2=8$ and $x^2-2y^2=4$

to prove they are orthogonal

solving the two curves, we have

$6y^2=4$

$y^2=\frac{2}{3}$

$y=\sqrt{\frac{2}{3}}$

$x=\pm \sqrt{\frac{16}{3}}$

Now, differentiating the two curve to get $m_1,m_2$-

Curve (1)- $2x+8y{y}^{\prime }=0$

$y^{\prime }=\frac{-2x}{8y}$

$m_1=\frac{-x}{4y}=-\sqrt{\frac{16}{3}}\times \frac{\sqrt{3}}{4\sqrt{2}}=\frac{-1}{\sqrt{2}}$

Curve (2)- $2x-4y{y}^{\prime }=0$

$y^{\prime }=\frac{2x}{4y}$

$m_2=\frac{x}{2y}=\sqrt{\frac{16}{3}}\times \frac{\sqrt{3}}{2\sqrt{2}}=\sqrt{2}$

$\therefore m_1{m}_2=\sqrt{2}\times \frac{-1}{\sqrt{2}}=-1$

$\theta =\frac{\pi }{2}$

Hence the curves are orthogonal.