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 $\theta$ is the acute angle between these curves at $P$, then $5 \sqrt{\tan \theta}=$

• A. 2
• B. $2 \sqrt{3}$
• C. $3 \sqrt{2}$
• D. 3

Answer 1

Answer: (C)

Points of intersection of curves
$y^{2} =32 x\,\,\,...(i)$
and $2 x^{2} =27 y\,\,\,...(ii)$
From Eqs. (i), and (ii), we get
$2 \cdot\left(\frac{y^{2}}{32}\right)^{2} =27 y$
$2 \cdot y^{4} =27 \cdot 32 \cdot 32 \cdot y $
$y =0, y^{3}=512 \cdot 27 $
$y =24$
From Eq. (i), we get
$x=0 $
$x=18$
So, coordinate of $P(18,24)$.
Equation of tangent through point $P(18,24)$ on the curve $y^{2}=32 x$
$\Rightarrow y \cdot 24 =16(x+18) $
$\Rightarrow 3 y =2 x+36$
$\therefore $ Slope $m_{1} =2 / 3$
Again equation of tangent through point $P(18,24)$ on the $2 x^{2}=27 y$
$\Rightarrow 2 x \cdot 18 =27 \frac{(y+24)}{2} $
$\Rightarrow 8 x =3 y+72$
$\Rightarrow 3 y =8 x-72$
Slope, $m_{2}=8 / 3$
Angle between these curve $\theta=$ Angle between tangents drawn from points $P$
$\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} $
So, $5 \sqrt{\tan \theta} =5 \cdot \sqrt{\frac{18}{25}} $
$=5 \cdot \frac{\sqrt{18}}{5}=3 \sqrt{2}.$