Points of intersection of curves


From Eqs. (i), and (ii), we get
$$
\begin{aligned}
2 \cdot\left(\frac{y^2}{32}\right)^2 & =27 y \\
2 \cdot y^4 & =27 \cdot 32 \cdot 32 \cdot y \\
y & =0, y^3=512 \cdot 27 \\
y & =24
\end{aligned}
$$

From Eq. (i), we get
$$
\begin{aligned}
& x=0 \\
& x=18
\end{aligned}
$$

So, coordinate of $P(18,24)$.
Equation of tangent through point $P(18,24)$ on the curve $y^2=32 x$
$$
\begin{aligned}
& \Rightarrow \quad y \cdot 24=16(x+18) \\
& \Rightarrow \quad 3 y=2 x+36 \\
& \therefore \quad \text { Slope } m_1=2 / 3 \\
&
\end{aligned}
$$

Again equation of tangent through point $P(18,24)$ on the $2 x^2=27 y$
$$
\begin{aligned}
\Rightarrow & 2 x \cdot 18 & =27 \frac{(y+24)}{2} \\
\Rightarrow & & 8 x=3 y+72 \\
\Rightarrow & 3 y & =8 x-72
\end{aligned}
$$

Slope, $m_2=8 / 3$
Angle between these curve $\theta=$ Angle between tangents drawn from points $P$
$$
\begin{aligned}
& \tan \theta=\frac{m_2-m_1}{1+m_1 m_2}=\frac{\frac{8}{3}-\frac{2}{3}}{1+\frac{8}{3} \cdot \frac{2}{3}} \\
& \tan \theta=\frac{6 / 3}{\frac{25}{9}}=\frac{18}{25}
\end{aligned}
$$

So,
$$
\begin{aligned}
5 \sqrt{\tan \theta} & =5 \cdot \sqrt{\frac{18}{25}} \\
& =5 \cdot \frac{\sqrt{18}}{5}=3 \sqrt{2} .
\end{aligned}
$$