The angle between the tangents to the curves $y=2 x^2$ and $x=2 y^2$ at $(1,1)$ is
The angle between the tangents to the curves $y=2 x^2$ and $x=2 y^2$ at $(1,1)$ is
$\tan ^{-1}\left(\frac{15}{8}\right)$
$\tan ^{-1}\left(\frac{7}{8}\right)$
$\tan ^{-1}\left(\frac{3}{4}\right)$
$\tan ^{-1}\left(\frac{1}{4}\right)$
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.
$\begin{aligned}
& \therefore \quad \tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right|=\left|\frac{4-\frac{1}{4}}{1+4 \times \frac{1}{4}}\right|=\frac{15}{8} \\
& \therefore \quad \theta=\tan ^{-1}\left(\frac{15}{8}\right)
\end{aligned}$