MCQ
The angle between the curves $y^2=x$ and $x^2=y$ at $(1,1)$ is:
  • A
    $\tan^{-1}\frac{4}{3}$
  • $\tan^{-1}\frac{3}{4}$
  • C
    $90^\circ$
  • D
    $45^\circ$

Answer

Correct option: B.
$\tan^{-1}\frac{3}{4}$
Given:
$\text{y}^2=\text{x} \ ...(1)$
$\text{x}^2=\text{y} \ ...(2)$
Point $= (1, 1)$
On diffierentiating $(1) \text{w.r.t. x},$ we get
$2\text{y}\frac{\text{dy}}{\text{dx}}=1$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{2\text{y}}$
$\Rightarrow\text{m}_1=\frac{1}{2}$
On differentiating $(2) \text{w.r.t. x},$ we get
$2\text{x}=\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\text{m}_2(1)=2$
Now,
$\tan\theta=\Big|\frac{\text{m}_1-\text{m}_2}{1+\text{m}_1\text{m}_2}\Big|=\Big|\frac{\frac{1}{2}-2}{1+\frac{1}{2}\times2}\Big|=\frac{3}{4}$
$\Rightarrow\theta=\tan^{-1}\big(\frac{3}{4}\big)$

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