# Thread: Angle between the curves

1. ## Angle between the curves

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°

2. Hello, metlx!

but it can be done more simply than you think . . .

what is the angle between: .$\displaystyle \begin{array}{cccc}x^2 + 4y^2 &=& 25 & {\color{blue}[1]} \\ 2x^2 + y^2 &=& 22 & {\color{blue}[2]} \end{array}$
$\displaystyle \begin{array}{cccc}\text{Multiply {\color{blue}[1]} by 2:} & 2x^2+8y^2&=&50 \\ \text{Subtract {\color{blue}[2]}:} & 2x^2 + y^2 &=& 22 \end{array}$

And we get: .$\displaystyle 7y^2 \:=\:28 \quad\Rightarrow\quad y^2 \:=\:4\quad\Rightarrow\quad y \:=\:\pm2$

Substitute into [1]: .$\displaystyle x^2 + 4(\pm2)^2 \:=\:25 \quad\Rightarrow\quad x^2 \:=\:9 \quad\Rightarrow\quad x \:=\:\pm3$

We have four intersections: .$\displaystyle A(3,2),\;B(\text{-}3,2),\;C(\text{-}3,\text{-}2),\;D(3,\text{-}2)$

We will use this theorem . . .

. . Given two lines with slopes $\displaystyle m_1$ and $\displaystyle m_2$,
. . the acute angle $\displaystyle \theta$ between them is given by: .$\displaystyle \tan\theta \:=\:\left|\frac{m_2 - m_1}{1 + m_1m_2}\right|$

Differentiate [1]: .$\displaystyle 2x + 8y\,\frac{dt}{dx} \:=\:0 \quad\Rightarrow\quad \frac{dy}{dx} \:=\:-\frac{x}{2y}$

Differentiate [2]: .$\displaystyle 4x + 2y\,\frac{dy}{dx} \:=\:0 \quad\Rightarrow\quad \frac{dy}{dx} \:=\:-\frac{2x}{y}$

At $\displaystyle A(3,2)\!:\;\;\begin{array}{ccccc}m_1 &=& \text{-}\frac{3}{2(2)} &=& \text{-}\frac{3}{4} \\ \\[-4mm] m_2 &=& \text{-}\frac{2(3)}{2} &=& \text{-}3 \end{array}$

Hence: .$\displaystyle \tan\theta_1 \:=\:\left|\frac{\text{-}3-(\text{-}\frac{3}{4})}{1 + (\text{-}\frac{3}{4})(\text{-}3)}\right| \;=\;\left|\frac{\text{-}\frac{9}{4}}{\frac{13}{4}}\right| \;=\;\frac{9}{13}$

. . Therefore: .$\displaystyle \theta_1 \;=\;\arctan\left(\tfrac{9}{13}\right) \;\approx\;34.7^o \quad\hdots\;\text{etc.}$

3. Differentiate [1]: .
I think you made a mistake there:
$\displaystyle \frac{dy}{dx} = -\frac {2x}{8y} = - \frac{x}{4y}$

so instead of -3/4 it's -3/8.

Thanks for explaining
It didn't occur to me to calculate the angles at each intersection..