The angle between the tangents to the curves | vedclass

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Question

(A) Given curves are $y=2x^2$ and $x=2y^2$.For the curve $y=2x^2$,the slope of the tangent is $\frac{dy}{dx} = 4x$.At the point $(1,1)$,the slope $m_1

Vedclass · Accepted Answer

$	an^{-1}\left(\frac{15}{8}ight)$

Vedclass · Answer

(A) Given curves are $y=2x^2$ and $x=2y^2$.For the curve $y=2x^2$,the slope of the tangent is $\frac{dy}{dx} = 4x$.At the point $(1,1)$,the slope $m_1 = 4(1) = 4$.For the curve $x=2y^2$,differentiating with respect to $x$,we get $1 = 4y \frac{dy}{dx}$,so $\frac{dy}{dx} = \frac{1}{4y}$.At the point $(1,1)$,the slope $m_2 = \frac{1}{4(1)} = \frac{1}{4}$.Let $	heta$ be the angle between the two tangents. The formula for the angle is $	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$.Substituting the values,$	an 	heta = \left| \frac{4 - 1/4}{1 + 4(1/4)} ight| = \left| \frac{15/4}{1 + 1} ight| = \frac{15/4}{2} = \frac{15}{8}$.Therefore,$	heta = 	an^{-1}\left(\frac{15}{8}ight)$.

The angle between the tangents to the curves $y=2x^2$ and $x=2y^2$ at $(1,1)$ is

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