Question

Statement I The angle of intersection of the curves $y^2=4 a x$ and $x^2=4 b y$ is
$\tan ^{-1}\left(\frac{3 a^{1 / 3} b^{1 / 3}}{2\left(a^{2 / 3}+b^{2 / 3}\right)}\right)$
Statement II $\tan \theta=\left|\frac{m_2-m_1}{1+m, m_2}\right|$, where $\theta$ is the angle between two curves whose slopes are $m_1$ and $m_2$.

• A. Statement I is true; statement II is true; statement II is a correct explanation for statement I
• B. Statement I is true; statement II is true; statement II is not a correct explanation for statement I
• C. Statement I is true; statement II is false
• D. Statement I is false, statement II is true

Answer 1

Answer: (A)

Given that, $ y^2=4 a x .....$(i)
and $ x^2=4 b y ......$(ii)
Solving Eqs. (i) and (ii), we get
$\left(\frac{x^2}{4 b}\right)^2 =4 a x$
$\Rightarrow x^4 =64 a b^2 x $
or $ x\left(x^3-64 a b^2\right) =0 $
$\Rightarrow x =0, x=4 a^{\frac{1}{3}} b^{\frac{2}{3}}$
Therefore, the points of intersection are $(0,0)$ and $\left(4 a^{\frac{1}{3}} b^{\frac{2}{3}}, 4 a^{\frac{2}{3}} b^{\frac{1}{3}}\right)$
Again, $ y^2=4 a x \Rightarrow \frac{d y}{d x}=\frac{4 a}{2 y}=\frac{2 a}{y}$
and $ x^2=4 b y$
$\Rightarrow \frac{d y}{d x}=\frac{2 x}{4 b}=\frac{x}{2 b}$
Therefore, at $(0,0)$ the tangent to the curve $y^2=4 a x$ is parallel to $Y$-axis and tangent to the curve $x^2=4 b y$ is parallel to $X$-axis.
$\Rightarrow$ Angle between curves $=\frac{\pi}{2}$.
At $\left(4 a^{\frac{1}{3}} b^{\frac{2}{3}}, 4 a^{\frac{2}{3}} b^{\frac{1}{3}}\right), m_1$ [slope of the tangent to the curve (i) $]=\frac{2 a}{4 a^{\frac{2}{3}} b^{\frac{1}{3}}}=\frac{1}{2}\left(\frac{a}{b}\right)^{\frac{1}{3}}$
$m_2$ [slope of the tangent to the curve (iii)] $-\frac{4 a^{\frac{1}{3}} b^{\frac{2}{3}}}{2 b}-2\left(\frac{a}{b}\right)^{\frac{1}{3}}$
If $Q$ is the angle between two curves
Therefore, $ \tan \theta =\left|\frac{m_2-m_1}{1+m_1 m_2}\right|=\left|\frac{2\left(\frac{a}{b}\right)^{\frac{1}{3}}-\frac{1}{2}\left(\frac{a}{b}\right)^{\frac{1}{3}}}{1+2\left(\frac{a}{b}\right)^{\frac{1}{3}} \frac{1}{2}\left(\frac{a}{b}\right)^{\frac{1}{3}}}\right| $
$ =\frac{3 a^{\frac{1}{3}} b^{\frac{1}{3}}}{2\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)}$
Hence, $ \theta =\tan ^{-1}\left[\frac{3 a^{\frac{1}{3}} \cdot b^{\frac{1}{3}}}{2\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)}\right]$