finding slope of the tangent by differentiating x and y with respect to t
\(\frac{dx}{dt}=2at\)
\(\frac{dx}{dt}=2a\)
Now dividing \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\) to obtain the slope of tangent
\(\frac{dy}{dx}=\frac{1}{t}\)
m(tangent) at t = 1 is 1
normal is perpendicular to tangent so, m1m2 = – 1
m(normal) at t = 1 is – 1
equation of tangent is given by y – y1 = m(tangent)(x – x1)
y – 2a = 1(x – a)
equation of normal is given by y – y1 = m(normal)(x – x1)
y – 2a = – 1(x – a)