Question

Statement I The angle of intersection of the curves $y^2=4ax$ and $x^2=4by$ is
${\tan }^{-1}\left(\frac{3a^{1/3}{b}^{1/3}}{2\left(a^{2/3}+b^{2/3}\right)}\right)$
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, $y^2=4ax.....$(i)
and $x^2=4by......$(ii)
Solving Eqs. (i) and (ii), we get
${\left(\frac{x^2}{4b}\right)}^2=4ax$
$\Rightarrow x^4=64a{b}^{2}x$
or $x\left(x^3-64a{b}^2\right)=0$
$\Rightarrow x=0,x=4a^{\frac{1}{3}}{b}^{\frac{2}{3}}$
Therefore, the points of intersection are $(0,0)$ and $\left(4a^{\frac{1}{3}}{b}^{\frac{2}{3}},4a^{\frac{2}{3}}{b}^{\frac{1}{3}}\right)$
Again, $y^2=4ax\Rightarrow \frac{dy}{dx}=\frac{4a}{2y}=\frac{2a}{y}$
and $x^2=4by$
$\Rightarrow \frac{dy}{dx}=\frac{2x}{4b}=\frac{x}{2b}$
Therefore, at $(0,0)$ the tangent to the curve $y^2=4ax$ is parallel to $Y$-axis and tangent to the curve $x^2=4by$ is parallel to $X$-axis.
$\Rightarrow$ Angle between curves $=\frac{\pi }{2}$.
At $\left(4a^{\frac{1}{3}}{b}^{\frac{2}{3}},4a^{\frac{2}{3}}{b}^{\frac{1}{3}}\right),m_1$ [slope of the tangent to the curve (i) $]=\frac{2a}{4a^{\frac{2}{3}}{b}^{\frac{1}{3}}}=\frac{1}{2}{\left(\frac{a}{b}\right)}^{\frac{1}{3}}$
$m_2$ [slope of the tangent to the curve (iii)] $-\frac{4a^{\frac{1}{3}}{b}^{\frac{2}{3}}}{2b}-2{\left(\frac{a}{b}\right)}^{\frac{1}{3}}$
If $Q$ is the angle between two curves
Therefore, $\tan \theta =\left|\frac{m_2-m_1}{1+m_1{m}_2}\right|=\left|\frac{2{\left(\frac{a}{b}\right)}^{\frac{1}{3}}-\frac{1}{2}{\left(\frac{a}{b}\right)}^{\frac{1}{3}}}{1+2{\left(\frac{a}{b}\right)}^{\frac{1}{3}}\frac{1}{2}{\left(\frac{a}{b}\right)}^{\frac{1}{3}}}\right|$
$=\frac{3a^{\frac{1}{3}}{b}^{\frac{1}{3}}}{2\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)}$
Hence, $\theta ={\tan }^{-1}\left[\frac{3a^{\frac{1}{3}}\cdot b^{\frac{1}{3}}}{2\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)}\right]$