show that the radius of curvature of the curve y^2=a^2(a-x)/x at (a,0)is a/2
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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.
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