y=f(x) and x=g(y) are two curves and P(x, y) | vedclass

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(C) Let $m_1$ be the slope of the tangent to the curve $y=f(x)$ at point $P$. Given $m_1 = \frac{dy}{dx} = Q(x)$.Let $m_2$ be the slope of the tangent

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tangent drawn at $P$ to one curve is normal to the other curve at $P$

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(C) Let $m_1$ be the slope of the tangent to the curve $y=f(x)$ at point $P$. Given $m_1 = \frac{dy}{dx} = Q(x)$.Let $m_2$ be the slope of the tangent to the curve $x=g(y)$ at point $P$. Given $\frac{dx}{dy} = -Q(x)$, we know that $m_2 = \frac{dy}{dx} = \frac{1}{dx/dy} = \frac{1}{-Q(x)}$.Since $m_1 	imes m_2 = Q(x) 	imes \left(-\frac{1}{Q(x)}ight) = -1$, the product of the slopes of the tangents at the point of intersection is $-1$.Therefore, the tangent drawn at $P$ to one curve is normal to the other curve at $P$.

$y=f(x)$ and $x=g(y)$ are two curves and $P(x, y)$ is a common point of the two curves. If at $P$, on the curve $y=f(x)$, $\frac{dy}{dx}=Q(x)$ and at the same point $P$ on the curve $x=g(y)$, $\frac{dx}{dy}=-Q(x)$, then

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