Question

Prove that parabola y²=x and hyperbola xy=k intersect at right angles, if 8k²=1.

Answer 1

Answer:

$\boxed{8k^2=1}$

Step-by-step explanation:

Prove that parabola y²=x and hyperbola xy=k intersect at right angles, if 8k²=1.

Given curves are

$y^2=x$......(1)

$xy=k$......(2)

First, we find the point of intersection of curves (1) and (2)

using (1) in (2)

$(y^2)y=k$

$y^3=k$

$y^3=k$

$y=k^{\frac{1}{3}}$

using the value of y in (1), we get

$x=k^{\frac{2}{3}}$

$\therefore\:\text{The point of intersection is }(k^{\frac{2}{3}}, k^{\frac{1}{3}})$

$y^2=x$

$\text{Differentiate with respect to x}$

$2y\:\frac{dy}{dx}=1$

$\frac{dy}{dx}=\frac{1}{2y}$

$\text{Slope of tangent }m_1=(\frac{dy}{dx})_(k^{\frac{2}{3}}, k^{\frac{1}{3}})=\frac{1}{2k^{\frac{1}{3}}}$

$xy=k$

$\text{Differentiate with respect to x}$

$x\:\frac{dy}{dx}+y.1=0$

$\implies\:\frac{dy}{dx}=\frac{-y}{x}$

$\text{Slope of tangent }m_1=(\frac{dy}{dx})_(k^{\frac{2}{3}}, k^{\frac{1}{3}})=\frac{-k^{\frac{1}{3}}}{k^{\frac{2}{3}}}$

$\text{since the curves are orthogonal, }\boxed{m_1\times\:m_2=-1}$

$\implies\:(\frac{1}{2k^{\frac{1}{3}}})\times(\frac{-k^{\frac{1}{3}}}{k^{\frac{2}{3}}})=-1$

$\implies\:(\frac{1}{2})\times(\frac{1}{k^{\frac{2}{3}}})=1$

$\implies\:\frac{1}{2k^{\frac{2}{3}}}=1$

$\implies\:2k^{\frac{2}{3}}=1$

$\text{Raising both sides to the power 3}$

$2^3(k^{\frac{2}{3}})^3=1^3$

$\boxed{8k^2=1}$