1. ## Parabola Trignometry

Hi, could someone please help me with part (a) of this question. I understand everything except that part...

Thanks heaps.

2. ## Re: Parabola Trignometry

Since it is in calculus i presume we can use differential.
equation of the parabola is
y^2 = 4px.
dy/dx = 2p/y
slope of tangent at the point ( x0, y0) will be = 2p/y0
That gives us the value of tan beta.

3. ## Re: Parabola Trignometry

Originally Posted by Vishak
Hi, could someone please help me with part (a) of this question. I understand everything except that part...
Note that $L$ is the tangent line to $y^2=4x$ at $(x_0,y_0)$.

$\beta$ is the angle $L$ makes with the positive $x$-axis. So the slope of $L$ is $\tan(\beta)$.

But that slope is $\\2yy'=4p\\y'=\frac{2p}{y}~.$

4. ## Re: Parabola Trignometry

Thanks heaps guys! Really appreciate it.