Question

angle between the curves x²=8y and y²=8x at (8,8) is
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Answer 1

Given Question :-

• Angle between the curves x²=8y and y²=8x at (8,8) is

$\huge \orange{AηsωeR} ✍$

Angle between the curves is defined as the angle between tangents to both the curves at their point of intersection.

☆ Let us consider

$\bf \: {x}^{2} = 8y$

☆Differentiate w. r. t. x, we get

$\bf \: ⟼ \dfrac{d}{dx} {x}^{2} = \dfrac{d}{dx} 8y$

$\bf \:  ⟼ 2x = 8\dfrac{dy}{dx}$

$\bf \: ⟼ \dfrac{dy}{dx} = \dfrac{x}{4}$

$\bf \:Slope \: of \: curve, \: at \: (8, 8) \: is$

$\bf\implies \:m_1 = \dfrac{dy}{dx}_{(8, 8)} = \dfrac{8}{4} = 2$

☆Now, Consider

$\bf \:  ⟼ {y}^{2} = 8x$

☆Differentiate w. r. t. x, we get

$\bf \: ⟼ \dfrac{d}{dx} {y}^{2} = \dfrac{d}{dx} 8x$

$\bf \: ⟼ 2y\dfrac{dy}{dx} = 8$

$\bf \: ⟼ \dfrac{dy}{dx} = \dfrac{4}{y}$

$\bf\implies \:Slope \: of \: curve, \: at \: (8, 8) \: is$

$\bf\implies \:m_2 = \dfrac{dy}{dx}_{(8, 8)} = \dfrac{4}{8} = \dfrac{1}{2}$

☆Let z be the angle between two curves, then

$\bf \:tanz = |\dfrac{m_1 - m_2}{1 + m_1 \times m_2} |$

$\sf \: On \: substituting \: the \: values \: of \: m_1 \: and \: m_2 \: we \: get$

$\bf \:  ⟼ tanz = |\dfrac{2 - \dfrac{1}{2} }{1 + 2 \times \dfrac{1}{2} } |$

$\bf \:  ⟼ tanz = |\dfrac{3}{4} |$

$\bf\implies \:z = {tan}^{ - 1} (\dfrac{3}{4} )$

${ \boxed {\bf{Hence \: angle \: between \: curves \: is \: {tan}^{ - 1} (\dfrac{3}{4} )}}}$