Question

The angle between the curves $\frac{x^2}{a^2+k_1}+\frac{y^2}{b^2+k_1}=1$ and $\frac{x^2}{a^2+k_2}+\frac{y^2}{b^2+k_2}=1$ is

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{2}$
• D. $ta{n}^{-1}\left(\frac{k_1}{k_2}\right)$

Answer 1

Answer: (C)

$\frac{\pi }{2}$

$\frac{dy}{dx}=-\frac{x(b^2+k_1)}{y(a^2+k_1)}=m_1$

$\frac{dy}{dx}=-\frac{x(b^2+k_2)}{y(a^2+k_2)}=m_2$

$m_2\times m_1=\frac{x^2}{y^2}\frac{(b^2+k_1)(b^2+k_2)}{(a^2+k_1)(a^2+k_2)}$ ---(1)

$\frac{x^2}{a^2+k_1}+\frac{y^2}{(b^2+k_2)}=\frac{x^2}{a^2+k_2}+\frac{y^2}{b^2+k^2}$

$\frac{x^2(k_2-k_1)}{(a^2+k_1)(a^2+k_2)}=\frac{y^2(k_1-k_2)}{(b^2+k_1)(b^2+k_2)}$ ----- (2)

$m_1\times m_2=-1$ (From eq (1) & (2))

$\therefore$ Angle b/w the curves is $\frac{\pi }{2}$