Question

The curves $\frac{x^2}{a^2+k_1}+\frac{y^2}{b^2+k_1}=1,\frac{x^2}{a^2+k_2}+\frac{y^2}{b^2+k_2}=1$ where $k_1\ne k_2$ intersect at an angle

• A. $0$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is D $\frac{\pi }{2}$

$\frac{x^2}{a^2+k_1}+\frac{y^2}{b^2+k_1}=1----\left(1\right)$ $\frac{x^2}{a^2+k_2}+\frac{y^2}{b^2+k_2}=1----\left(2\right)$

$m_1\Rightarrow \frac{dy}{dx}=-\left(\frac{b^2+k_1}{a^2+k_1}\right)\frac{x}{y}----\left(3\right)$ from $\left(1\right)$

$m_1\Rightarrow \frac{dy}{dx}=-\left(\frac{b^2+k_2}{a^2+k_2}\right)\frac{x}{y}----\left(4\right)$ from $\left(2\right)$

$m_1.m_2\Rightarrow \frac{(b^2+k_1)(b^2+k_2)}{(a^2+k_1)(a^2+k_2)}\frac{x^2}{y^2}----\left(5\right)$

from eq $\left(1\right)-$ eq $\left(2\right)$

$x^2(\frac{1}{a^2+k_1}-\frac{1}{a^2+k_2})+y^2(\frac{1}{b^2+k_1}-\frac{1}{b^2+k_1})=0$

$\frac{(b^2+k_1)(b^2+k_2)}{(a^2+k_1)(a^2+k_2)}=\frac{-y^2}{x^2}----\left(6\right)$

Putting value of eq $\left(6\right)$ in eq $\left(5\right)$

$m_1.m_2=-1$

Hence the angle between two curve is $\frac{\pi }{2}$.