Question

The angle of intersection of the curves $x^2+4y^2=32$ and $x^2-y^2=12$ at any point of their intersection is

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{2}$

Answer 1

Answer: (A)

$\frac{\pi }{6}$

$x^2+{4y}^2=32⟶\left(i\right)$

$x^2-y^2=12⟶\left(ii\right)$

Solving, $y=\pm 2$

$x=\pm 4$

$\therefore$ Point of $X^n$ are $(4,2),(4,-2),(-4,2),(-4,-2)$

At $(4,2)$

$m_1=\frac{-x}{4y}$ [differentiating $\left(i\right)$ wrt $x$]

$=\frac{2}{-16}=\frac{-1}{8}$

$m_2=y/x$ [differentiating $\left(ii\right)$ wrt $x$]

$=\frac{2}{4}=\frac{1}{2}$

$\tan \theta =\frac{|m_1-m_2|}{1+m_1{m}_2}=\frac{|\frac{-1}{8}-\frac{1}{2}|}{1-\frac{1}{16}}=\frac{\frac{2+8}{16}}{\frac{15}{16}}=\frac{10}{15}=\frac{2}{3}$

$\Rightarrow \theta ={\tan }^{-1}\left(2/3\right)\simeq \pi /6$ [A]