Question

Find the radius of curvature at any point (a cos theta, b sin theta) on ellipse x square /a square + y square / b square =1

Answer 1

The radius of curvature of given ellipse is $\frac{(a^2.sin^2.ϕ+b^2.cos^2.ϕ)^\frac{3}{2}}{a.b}$.

Given:

The given point, P = (a cos θ, b sin θ).

The equation of the ellipse = $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.$

To Find:

We have to find the radius of curvature at point P on the given ellipse.

Solution:

The tangent at the point P is given by,

$\frac{x}{a}.$ cos Φ + $\frac{y}{b}.$ sin Φ = 1.

The equation for the straight line passing through point, P is given by,

$\frac{cos Φ}{a} (x - a. cos Φ) - \frac{sin Φ}{b} (y - b.sin Φ) = 0$

Or, $\frac{x}{a}.cos Φ - \frac{y}{b} .sin Φ - cos 2Φ = 0.$

The equation for the circle of curvature is given by,

$\frac{x^2}{a^2} + \frac{y^2}{b^2} - 1 + λ[\frac{x}{a} .cos Φ + \frac{y}{b}.sin Φ - 1] [\frac{x}{a}.cos Φ - \frac{y}{b} .sin Φ - cos 2Φ] = 0.$

For a circle, the coefficients of x² and y² must be equal. Hence, the above equation becomes,

$\frac{1}{a^2} + λ . \frac{cos^2 Φ}{a^2} = \frac{1}{b^2} - λ . \frac{sin^2 Φ}{b^2}$

$λ = \frac{a^2 - b^2}{b^2. cos^2 Φ + a^2. sin^2 Φ}$.

On substituting the above equation for λ in the equation for the circle of curvature, we get,

$(b^2.cos^2 Φ + a^2. sin^2 Φ) (\frac{x^2}{a^2} + \frac{y^2}{b^2} - 1) + (a^2-b^2) [\frac{x^2}{a^2}.cos^2Φ - \frac{y^2}{b^2}.sin^2Φ- \frac{x.cosΦ}{a}.(1 + cos 2Φ) + \frac{y.cosΦ}{b}.(1 - cos 2Φ) +cos 2Φ ] = 0$

On simplifying the above equation, we get,

$[x - \frac{a^2 - b^2}{a}.cos^3 ]^2 + [y + \frac{a^2 - b^2}{b}.sin^3]^2 = \frac{a^2.sin^2Φ+b^2.cos^2Φ}{a^2.b^2}$

∴, The radius of curvature at the point P (a cos θ, b sin θ) = $\frac{(a^2.sin^2.ϕ+b^2.cos^2.ϕ)^\frac{3}{2}}{a.b}$.

Hence, the radius of curvature of given ellipse is $\frac{(a^2.sin^2.ϕ+b^2.cos^2.ϕ)^\frac{3}{2}}{a.b}$.

#SPJ2