# Math Help - Polar Curve Geometry

1. ## Polar Curve Geometry

Q: Sketch the curve with equation $r=a(1+\cos\theta)$ for $0\le\theta\le\pi$ where $a>0$. Sketch also the line with equation $r=2a\sec\theta$ for $-\frac{\pi}{2}<\theta<\frac{\pi}{2}$on the same diagram.
The half-line with equation $\theta=\alpha$, $0<\alpha<\frac{\pi}{2}$, meets the curve at $A$ and the line with equation $r=2a\sec\theta$ at $B$. If $O$ is the pole, find the value of $\cos\alpha$ for which $OB=2OA$.

I drew the sketch but can't do second part. Could someone help? Thanks in advance.

2. Originally Posted by Air
Q: Sketch the curve with equation $r=a(1+\cos\theta)$ for $0\le\theta\le\pi$ where $a>0$. Sketch also the line with equation $r=2a\sec\theta$ for $-\frac{\pi}{2}<\theta<\frac{\pi}{2}$on the same diagram.
The half-line with equation $\theta=\alpha$, $0<\alpha<\frac{\pi}{2}$, meets the curve at $A$ and the line with equation $r=2a\sec\theta$ at $B$. If $O$ is the pole, find the value of $\cos\alpha$ for which $OB=2OA$.

I drew the sketch but can't do second part. Could someone help? Thanks in advance.
We will see all points in $(r,\theta)$ format.

Clearly $A = (a(1+\cos \alpha),\alpha)$ and $B = (2a \sec \alpha, \alpha)$.

$OA = a(1+\cos \alpha)$ and $OB = 2a \sec \alpha$.

So $OB = 2OA \Rightarrow 2a \sec \alpha = 2a(1+\cos \alpha) \Rightarrow \cos^2 \alpha + \cos \alpha - 1 = 0$

P.S: Well I am not sure of it but it looks correct, if you have any doubts, just ask ok?It will help me too

3. Originally Posted by Isomorphism
We will see all points in $(r,\theta)$ format.

Clearly $A = (a(1+\cos \alpha),\alpha)$ and $B = (2a \sec \alpha, \alpha)$.

$OA = a(1+\cos \alpha)$ and $OB = 2a \sec \alpha$.

So $OB = 2OA \Rightarrow 2a \sec \alpha = 2a(1+\cos \alpha) \Rightarrow \cos^2 \alpha + \cos \alpha - 1 = 0$

P.S: Well I am not sure of it but it looks correct, if you have any doubts, just ask ok?It will help me too

How did you get the co-ordinates?

4. Originally Posted by Air
Q: Sketch the curve with equation $r=a(1+\cos\theta)$ for $0\le\theta\le\pi$ where $a>0$. Sketch also the line with equation $r=2a\sec\theta$ for $-\frac{\pi}{2}<\theta<\frac{\pi}{2}$on the same diagram.
The half-line with equation $\theta=\alpha$, $0<\alpha<\frac{\pi}{2}$, meets the curve at $A$ and the line with equation $r=2a\sec\theta$ at $B$. If $O$ is the pole, find the value of $\cos\alpha$ for which $OB=2OA$.
I drew the sketch but can't do second part. Could someone help? Thanks in advance.
Originally Posted by Air
How did you get the co-ordinates?
The question says : "The half-line with equation $\theta=\alpha$, $0<\alpha<\frac{\pi}{2}$, meets the curve at $A$"

So the point A where both the curves meet must satisfy both $\theta=\alpha$, $0<\alpha<\frac{\pi}{2}$ and $r=a(1+\cos\theta)$ for $0\le\theta\le\pi$ where $a>0$.

So substitute $\theta=\alpha$ in $r=a(1+\cos\theta)$, to get $r$. We already know that $\theta=\alpha$. Thus we can get $(r,\theta)$.

Do a similar thing for B by finding intersection with $r=2a\sec\theta$.

Now OA is the distance from the pole to the point A. That is nothing but the radial distance...