Find the equation of the normal line at the point H, where θ= π/6, on the curve with equations x=3sinθ and y=5cosθ

Step 1:

We begin solving this problem by finding the gradient of the tangent line at H, which is equal to dy/dx.

In this case, our curve has equations  x = 3sinθ and y= 5cosθ.

Now, we differentiate x and y with respect to θ

dx/dθ = 3cosθ 

dy/dθ = -5sinθ

Using the chain rule we get

dy/dx = dy/dθ * dθ/dx

dy/dx = -5sinθ/3cosθ

Now using that at the point H, θ= π/6

dy/dx = -5sin(π/6)/3cos(π/6)

dy/dx = -5(1/2)/3(3^1/2/2)

dy/dx = -5/3(3^1/2)

Step 2:

The gradient of the normal line is equal to - 1/M, where M is the gradient of the tangent line at the point H where θ= π/6.

Hence the gradient of the normal line is 3(3^1/2)/5

Step 3:

Now the general formula for the equation of a line at (X,Y) is y - Y = M(x - X)

In our case we have that, θ= π/6, hence

x=3sin(π/6)= 3/2    and

y=5cos(π/6)= 5(3^1/2)/2

Hence the equation of the normal line is,

y - 5(3^1/2)/2 = 3(3^1/2)/5(x - 3/2 )

y - 5(3^1/2)/2 =3(3^1/2)x/5 - 9(3^1/2)/10

Multiplying out every term by 10, we are left with

10y - 25(3^1/2) =6(3^1/2)x - 9(3^1/2)

10y=6(3^1/2)x + 16(3^1/2)

5y=3(3^1/2)x + 8(3^1/2)