# If the curve $y^{2}=x$ and $xy=k$ are orthogonal than prove that $8k^{2}=1.$

Toolbox:
• If $y=f(x)$ then $\large\frac{dy}{dx}$$=f'(x) is the rate of change of y w.r.t x • \large\frac{dy}{dx_{(x_1,y_1)}} is the slope of the tangent to the curve at the point (x_1,y_1) on the curve. It is the slope of the curve at that point. • The normal at a point (x_1,y_1) on y=f(x) is perpendicular to the tangent at (x_1,y_1) • If the curves y=f(x),y=g(x) intersect at (x_1,y_1) then the angle of intersection of the curves is the angle between the tangents at the point (x_1,y_1) • If m_1=t'(x_1) and m_2=g'(x_1) then the angle is \tan ^{-1} \large\frac{m_1-m_2}{1+m_1m_2} when the curves are orthogonal , the tangents are perpendicular to each other and m_1m_2=-1 Given y^2=x and xy=k are orthogonal Step 1: Let the curves intersect at (x_1,y_1) Therefore {y_1}^2=x_1----(i) and x_1y_1=k-----(ii) Substitute for y_1 from (ii) in (i) \large\frac{k^2}{{x_1}^2}$$=x_1=>{x_1}^3=k^2$-----(iii)
Step 2:
The slopes of the tangents of the curves $(m_1,m_2)$ at $(x_1,y_1)$ is $\large\frac{dy}{dx{(x_1y_1)}}$ for each curve
$y^2=x$
$2y \large\frac{dy}{dx}$$=1 \large\frac{dy}{dx}=\frac{1}{2y} Therefore m_1=\large\frac{1}{2y_1}=\frac{x_1}{2k} (from (ii)) xy=k y+x\large\frac{dy}{dx}$$=0=> \large\frac{dy}{dx}=\frac{-y}{x}$
$m_2=\large\frac{-y_1}{x_1}=\frac{-k}{{x_1}^2}$ (from (ii))
Step 3:
The curves are orthogonal
Therefore $m_1m_2=-1 =>\large\frac{x_1}{2k} \times \frac{-k}{{x_1}^2}$$=-1=>x_1=\large\frac{1}{2} Step 4: Substitute x_1=\large\frac{1}{2} in (iii) \large\frac{1}{8}$$=k^2 =>8h^2=1$