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%
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 % .....
$\displaystyle a \sec \alpha - a \sin \alpha \tan \alpha = \frac{a - a \sin^2 \alpha }{\cos \alpha } = \frac{a(1 - \sin^2 \alpha )}{\cos \alpha } = ....$
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:
$\displaystyle y - a \sec \alpha = \sin \alpha (x - a \tan \alpha )$
$\displaystyle \Rightarrow y - a \sec \alpha = \sin \alpha x - a \tan \alpha \sin \alpha$
$\displaystyle \Rightarrow y = \sin \alpha x - a \tan \alpha \sin \alpha + a \sec \alpha $.
And I've already all but shown you that $\displaystyle - a \tan \alpha \sin \alpha + a \sec \alpha$ simplifies to $\displaystyle a \cos \alpha$.