# Show that the normal at any point θ to the curve x = a cosθ + aθ sinθ, y = a sinθ - aθ cosθ is at a constant distance from the origin

**Solution:**

According to the given question

We have

x = a cos θ + aθ sin θ

Therefore,

dx/dθ = - a sin θ + a sin θ + aθ cos θ

= aθ cos θ

Also, y = a sin θ - aθ cos θ

Hence,

dy/dθ = a cos θ - a cos θ + aθ cos θ

= aθ cos θ

Thus,

dy/dx = (dy/dθ).(dθ/dx)

= aθ sin θ / aθ cos θ

= tan θ

Slope of the normal at any point θ is - 1/ tan θ

The equation of the normal at a given point ( x, y) is given by,

y - a sin θ + aθ cos θ = - 1/ tan θ(x - a cos θ - aθ sin θ)

⇒ y sin θ - a sin^{2} θ + aθ sin θ cos θ = - x cos θ + a cos^{2} θ + aθ sin θ cos θ

⇒ x cosθ + y sinθ - a (sin^{2} θ + cos^{2} θ) = 0

⇒ x cosθ + y sinθ - a = 0

Now, the perpendicular distance of the normal from the origin is

| -a | / cos^{2}θ sin^{2}θ

= |- a|/√1

= |- a|, which is independent of θ.

Hence, the perpendicular distance of the normal from the origin is constant

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

## Show that the normal at any point θ to the curve x = a cosθ + aθ sinθ, y = a sinθ - aθ cosθ is at a constant distance from the origin

**Summary:**

Hence we have shown that the normal at any point θ to the curve x = a cosθ + aθ sinθ, y = a sinθ - aθ cosθ is 'a' at a constant distance from the origin

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