Question

The curve given by the parametric equations;

Answer 1

Answer:

x=36, t = ±5/sqrt 3

Step-by-step explanation:

we are given a curve reprsented by parametric equations.

$x=36-t^2,\\ y=t^3-25t$

Also given that this is symmetric about x axis.

For finding a tangent we first find slope i.e. dy/dx

If dy/dx has dy/dt in numerator and dx/dt in the denominator

Hence for a tangent to be horizontal slope =0

i.e.dy/dt =0

Similarly for a tangent to be vertical slope=undefined

Or dx/dt =0

a) $\frac{dx}{dt} =-2t =0\\t=0\\x=36$

when x= 36, the tangent is horizontal

b) $\frac{dy}{dt} =0\\3t^2-25 =-\\t=\frac{5}{\sqrt{3} } ,-frac{5}{\sqrt{3} }$

For these t values tangent is vertical.