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# Find the slope of the normal to the curve x = 1 - a sinθ and y = b cos^{2} θ at θ = π/2

**Solution:**

The slope of a line is nothing but the change in y coordinate with respect to the change in x coordinate of that line

It is given that

x = 1 - a sin θ and y = b cos^{2} θ

Therefore,

dx/dθ

= d/dθ (1 - a sin θ)

= - a cos θ

dy/dθ

= d/dθ (b cos^{2} θ)

= - 2b sin θ cos θ

dy/dx = [(dy/dθ) / (dx/dθ)]

= (- 2b sinθ cosθ) / (- a cosθ)

= 2b/a sin θ

Now, the slope of the tangent at θ = π/2 is given by,

dy/dx]_{θ = π/2} = - tanθ]_{θ = π/2}

= 2b/a sin π/2

= 2b/a

Hence,

the slope of the normal at θ = π / 2 is given by,

- 1/slope of the tangent at θ = π / 2

= - 1/(2b/a)

= - a/2b

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

## Find the slope of the normal to the curve x = 1 - a sinθ and y = b cos^{2} θ at θ = π/2

**Summary:**

The slope of the normal to the curve x = 1 - a sinθ and y = b cos^{2} θ at θ = π/2 is - a/2b. The slope of a line is nothing but the change in y coordinate with respect to the change in x coordinate of that line

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