Question

Find whether the curves x^2 +y^2=8 and x^2-2y^2=4 orthogonal

Answer 1

Two curves are said to be orthogonal if product of their slopes is -1.

The curve 1 is ,

x² +y²=8-----(A)

Differentiating once with respect to x,

2 x + 2 y y'=0

x + y y'=0

$y'=\frac{-x}{y}$-----(1)

The curve 2 is ,

x² - 2 y²=4------(B)

Differentiating once with respect to x,

2 x - 4 y y'=0

x- 2 y y'=0

$Y'=\frac{x}{2 y}$-------(2)

Equation A - Equation B

$3y^2=4\\\\ y=\frac{\pm2}{\sqrt{3}}$

2 Equation A + Equation B,

$3x^2=20\\\\x=\frac{2\sqrt{5}}{\sqrt{3}}$

Product of slopes of curve (1) and curve (2) is

y' Y' $=\frac{-x}{y}\times \frac{x}{2y}=\frac{-x^2}{2y^2}$

Substituting the value of x, and y in above product of slopes,

$y'Y'=\frac{\frac{-20}{3}}{2*\frac{4}{3}}=\frac{-5}{2}$

As, product of slopes of two curves at point $(\frac{2\sqrt{5}}{\sqrt{3}},\frac{\pm2}{\sqrt{3}})$ is not equal to -1, so the two curves are not orthogonal.