Find the slope of the normal to the curve x = 1 - a sinθ and y = b cos2 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 cos2 θ
Therefore,
dx/dθ
= d/dθ (1 - a sin θ)
= - a cos θ
dy/dθ
= d/dθ (b cos2 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 cos2 θ at θ = π/2
Summary:
The slope of the normal to the curve x = 1 - a sinθ and y = b cos2 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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