Q61 / 150 · MHT CET 2023 (12 May Shift 1) · Mathematics

The angle between the tangents to the curves $y=2 x^2$ and $x=2 y^2$ at $(1,1)$ is

  • Correct answer
    A
    $\tan ^{-1}\left(\frac{15}{8}\right)$
  • B
    $\tan ^{-1}\left(\frac{7}{8}\right)$
  • C
    $\tan ^{-1}\left(\frac{3}{4}\right)$
  • D
    $\tan ^{-1}\left(\frac{1}{4}\right)$

Solution

$$ y=2 x^2 $$ $\therefore \quad$ Slope of the tangent to this curve is $\frac{\mathrm{d} y}{\mathrm{~d} x}=\mathrm{m}_1=4 x$ $\therefore \quad$ at $(1,1), \mathrm{m}_1=4$ $$ x=2 y^2 $$ $\therefore \quad$ Slope of the tangent to this curve is $\frac{\mathrm{d} y}{\mathrm{~d} x}=\mathrm{m}_2=\frac{1}{4 y}$ $\therefore \quad$ at $(1,1), \mathrm{m}_2=\frac{1}{4}$ Let $\theta$ be the angle between two tangents. $$ \begin{aligned} & \therefore \quad \tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right|=\left|\frac{4-\frac{1}{4}}{1+4 \times \frac{1}{4}}\right|=\frac{15}{8} \\ & \therefore \quad \theta=\tan ^{-1}\left(\frac{15}{8}\right) \end{aligned} $$

Hence, the correct answer is option: 'A'.

Relevant Questions