what is the angle between:

$\displaystyle x^2 + 4y^2 = 25$

and

$\displaystyle 2x^2 + y^2 = 22$

Before I explain what causes me problems I'll show how far I managed to get..

1.The coordinates when the curves cross

from the first equation: $\displaystyle y = \sqrt{\frac{25-x^2}{4}} $

from the second equation: $\displaystyle y = \sqrt{22-2x^2}$

$\displaystyle \frac{25-x^2}{4} = 22 - 2x^2 $

$\displaystyle 7x^2 = 63$

$\displaystyle x_1 = 3 \Rightarrow y_1 = 2 , y_2 = -2$

$\displaystyle x_2 = -3 \Rightarrow y_3 = 2 , y_4 = -2$

I have 4 points..

$\displaystyle T_1 (3,2)$

$\displaystyle T_2 (-3,2)$

$\displaystyle T_3 (-3,2)$

$\displaystyle T_4 (-3,-2)$

2. The derivative of both curves

First curve: $\displaystyle (x^2 + 4y^2 = 25)' $

$\displaystyle y' = \frac{-x}{4y} = \frac{-x}{4\sqrt{\frac{25-x^2}{4}}} $

$\displaystyle f'_1(3) = -\frac{3}{8} =k_1$

$\displaystyle f'_2(3) = \frac{3}{8} = k_2$

$\displaystyle f'_1(-3) = \frac{3}{8} =k_3$

$\displaystyle f'_2(-3) = -\frac{3}{8} = k_4$

Second curve: $\displaystyle 2x^2 + y^2 = 22$

$\displaystyle y' = \frac {-2x}{y} = \frac{-2x}{\sqrt{22-2x^2}} $

$\displaystyle f'_1(3) = -3 =k_5$

$\displaystyle f'_2(3) = 3 = k_6$

$\displaystyle f'_1(-3) = 3 =k_7$

$\displaystyle f'_2(-3) = -3 = k_8$

Ok I'm not sure what to do now.. Was it even right to make out 8 k's?

after all.. k1 = k4, k2 = k3, k5 =k8, k6 = k7

I know that I have to use the following equation:

$\displaystyle tan \phi = |\frac{k_1 - k_2}{1 + k_1k_2}|$

but I dunno which k's to put into equation..

The answer is supposed to be 51°