The angle of intersection between curves $y = x^{3} a n d 6 y = 7 - x^{2}$ at point $(1, 1)$ is -

\frac{π}{4}

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\frac{π}{3}

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\frac{π}{2}

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None of these

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Solution

The correct option is C $\frac{π}{2}$
Let $c_{1} : y = x^{3}$ and $c_{2} : 6 y = 7 - x^{2}$
Differentiating w.r.t $x$
$c_{1} : \frac{d y}{d x} = 3 x^{2}$ and $c_{2} : 6 \frac{d y}{d x} = - 2 x$
${(\frac{d y}{d x})}_{c_{1}} = 3 x^{2}$ and ${(\frac{d y}{d x})}_{c_{2}} = - \frac{x}{3}$
Thus slope of tangent at $(1, 1)$ to the given curve are $m_{1} = 3$ and $m_{2} = - \frac{1}{3}$
Therefore, angle of intersection between the given curve is, $θ = {tan}^{- 1} ∣ ∣ ∣ \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} ∣ ∣ ∣ = \frac{π}{2}$
Hence, option 'C' is correct.

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The angle of intersection between curves y=x3and6y=7−x2 at point (1,1) is -

The angle of intersection between curves $y = x^{3} a n d 6 y = 7 - x^{2}$ at point $(1, 1)$ is -