Find the slope of the normal to the curve x = a cos³ θ, y = a sin³ θ at θ = π/4

Solution:

For a curve y = f(x) containing the point (x₁,y₁) the equation of the tangent line to the curve at (x₁,y₁) is given by 

y − y₁ = f′(x₁) (x − x₁).

The given curve is

x = a cos³ θ, y = a sin³ θ

Therefore,

dx/dθ

= d/dθ (a cos³ θ)

= - 3a cos² θ sin θ

dy/dθ

= d/dθ (a sin³ θ)

= 3a sin² θ (cos θ)

dy/dx =

[(dy/dθ) / (dx/dθ)]

= (- 3a cos² θ sin θ) / (3a sin² θ (cos θ))

= - sinθ/cosθ

= - tanθ

Now, the slope of the tangent to a curve at θ = π / 4 is given by,

dy/dx]_{θ = π/4} = - tanθ]_{θ = π/4}

= - tan π / 4

= - 1

Hence, the slope of the normal at θ = π / 4 is given by,

- 1/slope of the tangent at θ = π / 4

= - 1 / - 1 = 1

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NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.3 Question 5

Find the slope of the normal to the curve x = a cos³ θ, y = a sin³ θ at θ = π/4.

Summary:

The slope of the normal to the curve x = a cos³ θ, y = a sin³ θ at θ = π/4 is 1. The slope of a line is nothing but the change in y coordinate with respect to the change in x coordinate of that line