The acute angle between the tangents drawn at | vedclass

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(B) Given curves are $x^2=4y$ $(i)$ and $y^2=4x$ $(ii)$.To find the point of intersection,substitute $y = \frac{x^2}{4}$ into $(ii)$:$\left(\frac{x^2}

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$\sin^{-1}\left(\frac{3}{5}\right)$

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(B) Given curves are $x^2=4y$ $(i)$ and $y^2=4x$ $(ii)$.To find the point of intersection,substitute $y = \frac{x^2}{4}$ into $(ii)$:$\left(\frac{x^2}{4}ight)^2 = 4x \implies \frac{x^4}{16} = 4x \implies x^4 = 64x \implies x(x^3 - 64) = 0$.So,$x=0$ or $x=4$. For $x=4$,$y = \frac{16}{4} = 4$. Thus,the point of intersection other than the origin is $(4,4)$.Now,differentiate $(i)$ with respect to $x$: $2x = 4 \frac{dy}{dx} \implies \frac{dy}{dx} = \frac{x}{2}$.At $(4,4)$,the slope $m_1 = \frac{4}{2} = 2$.Differentiate $(ii)$ with respect to $x$: $2y \frac{dy}{dx} = 4 \implies \frac{dy}{dx} = \frac{2}{y}$.At $(4,4)$,the slope $m_2 = \frac{2}{4} = \frac{1}{2}$.The angle $	heta$ between the tangents is given by $	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$.$	an 	heta = \left| \frac{2 - \frac{1}{2}}{1 + 2 	imes \frac{1}{2}} ight| = \left| \frac{\frac{3}{2}}{2} ight| = \frac{3}{4}$.Since $	an 	heta = \frac{3}{4}$,we have $\sin 	heta = \frac{3}{5}$,so $	heta = \sin^{-1}\left(\frac{3}{5}ight)$.

The acute angle between the tangents drawn at the point of intersection (other than the origin) of the curves $x^2=4y$ and $y^2=4x$ is

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