1. ## Parametric Equation (hard)

am going to use % as alpha sign

the parametric equations of the curve C are:

x = a tan %
y = a sec %

dy/dx = sin%

show that the equation of the tangent to C at pont P(a tan %, a sec %), where 0<%<(pie/2) is

y = x sin% + a cos%

2. Originally Posted by Stylis10
am going to use % as alpha sign

the parametric equations of the curve C are:

x = a tan %
y = a sec %

dy/dx = sin%

show that the equation of the tangent to C at pont P(a tan %, a sec %), where 0<%<(pie/2) is

y = x sin% + a cos%
Use y - y1 = m(x - x1) as your model for the tangent line.

Sub the known stuff in.

Done that? Then you have to simplify a sec % - a sin % tan % and get a cos % .....

$a \sec \alpha - a \sin \alpha \tan \alpha = \frac{a - a \sin^2 \alpha }{\cos \alpha } = \frac{a(1 - \sin^2 \alpha )}{\cos \alpha } = ....$

3. im still not getting it

4. Originally Posted by mr fantastic
Use y - y1 = m(x - x1) as your model for the tangent line.

Sub the known stuff in.

Done that? Then you have to simplify a sec % - a sin % tan % and get a cos % .....

$a \sec \alpha - a \sin \alpha \tan \alpha = \frac{a - a \sin^2 \alpha }{\cos \alpha } = \frac{a(1 - \sin^2 \alpha )}{\cos \alpha } = ....$
y - y1 = m(x - x1) is the standard slope-point form of a line.

m is the gradient of the line. (x1, y1) is a known point on the line.

You're given a point on the tangent, so x1 and y1 are known. And you've got dy/dx = m. So sub the stuff you know into the above:

$y - a \sec \alpha = \sin \alpha (x - a \tan \alpha )$

$\Rightarrow y - a \sec \alpha = \sin \alpha x - a \tan \alpha \sin \alpha$

$\Rightarrow y = \sin \alpha x - a \tan \alpha \sin \alpha + a \sec \alpha$.

And I've already all but shown you that $- a \tan \alpha \sin \alpha + a \sec \alpha$ simplifies to $a \cos \alpha$.

5. oh thnxs, i get it now, i was trying to simplify sin% to get a numerical value for m.