Question

$\huge \tt{Question}$

Find the point on the curve y = x³ - 3x where the tangent is parallel to the chord joining the points (1, -2) and (2, 2). ​

Answer 1

★ $\large{\underline{\rm{\green{Solution:-}}}}$

Here, y = (x³ - 3x)

$\implies \rm \frac{dy}{dx} = \frac{d}{dx} \: ( {x}^{3} - 3x)$

$\implies {\boxed{ \rm {= 3 {x}^{2} - 3}}}$

Therefore, slope of tangent = 3x³ - 3.

★ Now, slope of the chord joining points (1, -2) and (2, 2) is

$\implies \rm\frac{2 - ( - 2)}{2 - 1} = \frac{4}{1} = 4$

★ Since tangent is parallel to the chord joining (1, -2) and (2, 2).

$\therefore \: \rm 3 {x}^{2} = 4 \: \implies \: {x}^{2} = \frac{7}{3} \: \implies x = \pm \sqrt{ \frac{7}{3} }$

★ $\boxed{\rm When \: \: x = \sqrt{ \frac{7}{3} }}$

$\implies \rm Then \: \: \: y = ( \sqrt{ \frac{7}{3} })^3 - 3. \sqrt{ \frac{7}{3} } = \frac{7}{3} \sqrt{ \frac{7}{3} } - 3 \sqrt{ \frac{7}{3} } = - \frac{2}{3} \sqrt{ \frac{7}{3} }$

★ $\boxed{\rm When \: \: x = - \sqrt{ \frac{7}{3} }}$

$\implies \rm Then \: \: \: y = ( - \sqrt{ \frac{7}{3} })^3 - 3( - \sqrt{ \frac{7}{3} } ) = - \frac{7}{3} \sqrt{ \frac{7}{3} } + 3 \sqrt{ \frac{7}{3} } = \frac{2}{3} \sqrt{ \frac{7}{3} }$

$\rm \: Thus \: required \: points \: are \: (\sqrt{ \frac{7}{3} },- \frac{2}{3} \sqrt{ \frac{7}{3} } \: and \: ( - \sqrt{ \frac{7}{3} }, \frac{2}{3} \sqrt{ \frac{7}{3} } )$

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