If $P$ and the origin are the points of intersection of the parabolas $y^2=32 x$ and $2 x^2=27 y$; and if $ heta$ is the acute angle between these curves at $P$, then $5 \sqrt{ an heta}=$

Question

If $P$ and the origin are the points of intersection of the parabolas $y^2=32 x$ and $2 x^2=27 y$; and if $	heta$ is the acute angle between these curves at $P$, then $5 \sqrt{	an 	heta}=$

Aicharya · Accepted Answer

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} $

If $P$ and the origin are the points of intersection of the parabolas $y^2=32 x$ and $2 x^2=27 y$; and if…

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