Find the equation of the tangent and the normal to the following curves at the indicated points: x - 2 at^2/1+t^2, y= 2at^3/1+t^2 at t = 1/2

Question

Find the equation of the tangent and the normal to the following curves at the indicated points:

\(x=\frac{2at^2}{1+t^2},\)\(y=\frac{2at^3}{1+t^2}\) at t = 1/2

Answer 1

finding slope of the tangent by differentiating x and y with respect to t

\(\frac{dx}{dt}\)\(=\frac{(1+t^2)4at-2at^2(2t)}{(1+t^2)^2}\)

\(\frac{dx}{dt}\)\(=\frac{4at}{(1+t^2)^2}\)

\(\frac{dy}{dt}\)\(=\frac{(1+t^2)6at^2-2at^3(2t)}{(1+t^2)^2}\)

\(\frac{dy}{dt}\)\(=\frac{6at^2+2at^4}{(1+t^2)^2}\)

Now dividing \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) to obtain the slope of tangent

\(\frac{dy}{dx}=\frac{6at^2+2at^4}{4at}\)

m(tangent) at t = \(\frac{1}{2}\) is \(\frac{13}{16}\)

normal is perpendicular to tangent so, m₁m₂ = – 1

m(normal) at t = \(\frac{1}{2}\) is \(-\frac{16}{13}\)

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

\(y-\frac{a}{5}=\frac{13}{16}(x-\frac{2a}{5})\)

equation of normal is given by y – y₁ = m(normal)(x – x₁)

\(y-\frac{a}{5}=-\frac{16}{13}(x-\frac{2a}{5})\)