The angle of intersection between the curves y = 4 - x^2 and y = x^2 is : Option 1: ()/2 Option 2: ()/4 Option 3: ^(-1) 4/3 Option 4: ^(-1) 4√2/7

Answer: ^(-1) 4√2/7. Solution: Solving 4-x^2=x^2 gives x=±√2. At x=√2: slope of y=4-x^2 is -2√2, slope of y=x^2 is 2√2. =, (-2√2-2√2)/(1+(-2√2)(2√2)), =4√2/7. So =^(-1)4√2/7.

Calculus

Application of Derivatives

Medium

common

The angle of intersection between the curves $y = 4 - x^2$ and $y = x^2$ is :

$\frac{\pi}{2}$

$\frac{\pi}{4}$

$\tan^{-1} \frac{4}{3}$

$\tan^{-1} \frac{4\sqrt{2}}{7}$

✅ Correct Option: 4

Solving $4-x^2=x^2$ gives $x=\pm\sqrt{2}$ .

At $x=\sqrt{2}$ : slope of $y=4-x^2$ is $-2\sqrt{2}$ , slope of $y=x^2$ is $2\sqrt{2}$ .

$\tan\theta=\left|\frac{-2\sqrt{2}-2\sqrt{2}}{1+(-2\sqrt{2})(2\sqrt{2})}\right|=\frac{4\sqrt{2}}{7}$ .

So $\theta=\tan^{-1}\frac{4\sqrt{2}}{7}$ .

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The angle of intersection between the curves $y = 4 - x^2$ and $y = x^2$ is :