Conics - Equation of normal to the hyperbola parametric form (1 Viewer)

[BobMac]

I'm able to derive the:
Normal and Tangent to the ellipse both parametric and Cartesian form.
Normal and tangent to the hyperbola Cartesian form and parametric form for the tangent. But I am unable to derive the equation of the normal to the hyperbola in the parametric form.
The answer is {[ax/sec] + [by/tan] = 1}
For some reason I can't seem to derive this. All the rest are fine except this. Any help would be greatly appreciated.
Thanks in advanced

 

[Shadowdude]

Since you can derive the parametric form of the tangent for the hyperbola - all you need is this: (gradient of normal to hyperbola) = -1 / (gradient of tangent to hyperbola).

Then you use your point-gradient form, taking care to use the parametric forms of x and y, and it should pop right out.

What are you having trouble with anyway? There's probably a trigonometric identity you're unaware of - but note that the hyperbola has sec and tan as its parameters - and 1 + [tan(x)]^2 = [sec(x)]^2