Question

Prove that radius of curvature is y^2/2 of the curve y = a/2 (e^a/x +e^-a/x)

Answer 1

Answer:

The equation of the curve, (a – x)y2 = (a + x)x2 passes through the origin. To see the nature of the tangent at the origin, equate to zero the lowest degree terms in x and y, i.e.

ay2 = ax2 or y = ± x

i.e., at the origin, neither of the axis are tangent to the given curve

∴ Putting

or

On comparing the coefficients of x2 and x3, we get

ap2 = a ⇒ p = ± 1

and apq – p2 = 1 ⇒ q = ± 2/ a

∴

Hence ρ(0, 0) is numerically a√2.

Alternately:

Equation of the curve is

∴

i.e.,

so that

and

At (0, 0) y1 = ± 1, y2 = ± 2/a

∴

Hence ρ(0, 0) is numerically a√2