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

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  • If $y=f(x)$ then $\large\frac{dy}{dx}$$=f'(x)$ is the rate of change of $y$ w.r.t $x$
  • $\large\frac{dy}{dx_{(x_1,y_1)}}$ is the slope of the tangent to the curve at the point $(x_1,y_1) $ on the curve. It is the slope of the curve at that point.
  • The normal at a point $(x_1,y_1)$ on $y=f(x)$ is perpendicular to the tangent at $(x_1,y_1)$
  • If the curves $y=f(x),y=g(x)$ intersect at $(x_1,y_1)$ then the angle of intersection of the curves is the angle between the tangents at the point $(x_1,y_1)$
  • If $m_1=t'(x_1)$ and $m_2=g'(x_1)$ then the angle is $\tan ^{-1} \large\frac{m_1-m_2}{1+m_1m_2}$ when the curves are orthogonal , the tangents are perpendicular to each other and $m_1m_2=-1$
Given $y^2=x$ and $xy=k $ are orthogonal
Step 1:
Let the curves intersect at $ (x_1,y_1) $
Therefore ${y_1}^2=x_1$----(i) and $x_1y_1=k$-----(ii)
Substitute for $y_1$ from (ii) in (i)
Step 2:
The slopes of the tangents of the curves $(m_1,m_2)$ at $(x_1,y_1)$ is $\large\frac{dy}{dx{(x_1y_1)}}$ for each curve
$2y \large\frac{dy}{dx}$$=1$
Therefore $m_1=\large\frac{1}{2y_1}=\frac{x_1}{2k}$ (from (ii))
$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$
answered Jul 24, 2013 by meena.p

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