Question

find the equation of tangent and normal of curve x=cost,y=sint at t=45degrees​

Answer 1

To find equations of tangent and normal to of a curve at the points (cos t, sin t).

We know that,

$\sf \: slope = m = \dfrac{dy}{dx}$

The above derivative is in parametric form (multiplying and dividing by dt).

$\implies \sf \: m = \dfrac{dy}{dx} = \dfrac{d(sin \: t)}{dt} \times \dfrac{dt}{d(cos \: t)} \\ \\ \implies \sf \: m = \dfrac{dy}{dx} = - cot \: t \\ \ \implies \sf \: m_{tangent} = \dfrac{dy}{dx} \bigg| _{t = \frac{\pi}{4}} = - 1$

Slope of the tangent is - 1.

At (x,y) = (cos t, sin t) and t = π/4, the points would be (1/√2,1/√2).

Using point slope form,

$\implies \sf y - \dfrac{1}{ \sqrt{2} } = - 1(x - \dfrac{1}{ \sqrt{2} } ) \\ \\ \implies \sf \sqrt{2} y - 1 = - \sqrt{2x} + 1 \\ \\ \implies \sf \: \sqrt{2}x + \sqrt{2} y = 2 \\ \\ \implies \boxed{ \boxed{\sf \: x + y = \sqrt{2} }}$

Equation of the tangent is x + y = √2.

We know that,

$\sf m_{normal} = - \dfrac{1}{m_{tangent}} \\ \\ \longrightarrow \sf \: m = 1$

At (x,y) = (cos t, sin t) and t = π/4, the points would be (1/√2,1/√2)

Using point slope form,

$\implies \sf y - \dfrac{1}{ \sqrt{2} } = 1(x - \dfrac{1}{ \sqrt{2} } ) \\ \\ \implies \sf \sqrt{2} y - 1 = \sqrt{2x} - 1 \\ \\ \implies \sf \: \sqrt{2}x - \sqrt{2} y = 0 \\ \\ \implies \boxed{ \boxed{\sf \: x - y = 0 }}$

Equation of the normal is x - y = 0.