Find the equation of the tangent and the normal to the following curve

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

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»

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)

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