The angle between the curves $y = 2 x$ and $x^{3} + y^{3} = 3 x y$ at the point other than origin is

\frac{π}{6}

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\frac{π}{3}

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\frac{π}{2}

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t a n^{- 1} (\frac{6}{13})

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Solution

The correct option is A $t a n^{- 1} (\frac{6}{13})$
$y^{'} = 2$
$3 x^{2} + 3 y^{2} y^{1} = 3 (y + x y^{1})$
$y^{1} (y^{2} - x) = 3 (y + x^{2})$
$y^{'} = \frac{(y - x^{2})}{y^{2} - x}$ --------(1)
$x^{3} + 8 x^{3} = 6 x^{2}$
$9 x^{3} = 6 x^{2}$
$(x = 2 / 3)$
$y = 4 /_{3}$
$y^{'} = \frac{(\frac{4}{3} - \frac{4}{9})}{(\frac{16}{9} - \frac{2}{3})}$
$= \frac{\frac{4 (3 - 1)}{9}}{(\frac{16 - 6}{9})}$
$= \frac{8}{10}$
$= \frac{4}{5}$
$t a n θ = \frac{2 - \frac{4}{5}}{1 + 2 \times \frac{4}{5}}$
$t a n θ = \frac{6}{13}$
$θ = t a n^{- 1} (\frac{6}{13})$

Suggest Corrections

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