Find the equation of the tangent and the normal to the curves at the i

1.	Find the equation of the tangent and the normal to the curves at the indicated points: x = a sec t, y = b tan t at t.
Answer» Given as x = a sec t, y = b tan t at t Differentiate with respect to x, to get the slope of tangent dx/dt = a sec t tan t dy/dt = b sec² t Dividing dy/dt and dx/dt to obtain the slope of tangent dy/dx = b cosec t/a m(tangent) at t = b cosec t/a The normal is perpendicular to tangent therefore, m₁m₂ = – 1 m(normal) at t = (-a/b)sin t The equation of tangent is given by y – y₁ = m(tangent)(x – x₁) y - b tan t = (b cosec t/a)(x - a sec t) The equation of normal is given by y – y₁ = m(normal)(x – x₁) y - b tan t = (-a sin t/b)(x - a sec t)

Find the equation of the tangent and the normal to the curves at the indicated points: x = a sec t, y = b tan t at t.

Answer»

Given as x = a sec t, y = b tan t at t

Differentiate with respect to x, to get the slope of tangent

dx/dt = a sec t tan t

dy/dt = b sec² t

Dividing dy/dt and dx/dt to obtain the slope of tangent

dy/dx = b cosec t/a

m(tangent) at t = b cosec t/a

The normal is perpendicular to tangent therefore, m₁m₂ = – 1

m(normal) at t = (-a/b)sin t

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

y - b tan t = (b cosec t/a)(x - a sec t)

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

y - b tan t = (-a sin t/b)(x - a sec t)