The angle between the curves y^2=4x+4 and y^2 | vedclass

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(D) Given curves are $y^2=4x+4$ $(i)$ and $y^2=36(9-x)$ (ii).To find the intersection points,equate the two expressions for $y^2$:$4x+4 = 324-36x$$40x

Vedclass · Accepted Answer

$90$

Vedclass · Answer

(D) Given curves are $y^2=4x+4$ $(i)$ and $y^2=36(9-x)$ (ii).To find the intersection points,equate the two expressions for $y^2$:$4x+4 = 324-36x$$40x = 320 \Rightarrow x = 8$.Substituting $x=8$ into $(i)$,$y^2 = 4(8)+4 = 36 \Rightarrow y = \pm 6$.So,the intersection points are $(8,6)$ and $(8,-6)$.Differentiating $(i)$ with respect to $x$: $2y \frac{dy}{dx} = 4 \Rightarrow \frac{dy}{dx} = \frac{2}{y}$.Differentiating (ii) with respect to $x$: $2y \frac{dy}{dx} = -36 \Rightarrow \frac{dy}{dx} = \frac{-18}{y}$.At point $(8,6)$:$m_1 = \frac{2}{6} = \frac{1}{3}$ and $m_2 = \frac{-18}{6} = -3$.Since $m_1 	imes m_2 = \frac{1}{3} 	imes (-3) = -1$,the tangents are perpendicular.Therefore,the angle between the curves is $90^{\circ}$ or $\frac{\pi}{2}$.

The angle between the curves $y^2=4x+4$ and $y^2=36(9-x)$ is (in $^{\circ}$)

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