Question

prove that evolute of ellipse of the curve 2xy=a^2 is (x+y)^2/3-(x-y)^2/3 =2a^2/3​

Answer 1

Step-by-step explanation:

The given curve is

2xy = a²

or, xy = a²/2 = b² (say) ..... (1)

Differentiating both sides with respect to x, we get

x dy/dx + y = 0 ..... (2)

Again differentiating both sides of (2) with respective to x, we get

x d²y/dx² + dy/dx + dy/dx = 0

or, x d²y/dx² + 2dy/dx = 0

or, x d²y/dx² + 2 (- y/x) = 0, by (2)

or, d²y/dx² = 2y/x²

Thus we have

dy/dx = - y/x

d²y/dx² = 2y/x²

If (X, Y) be the coordinates of the centre of curvature, then

X = x - y₁ (1 + y₁²)/y₂

= x - (- y/x) (1 + y²/x²)/(2y/x²)

= x + x (1 + y²/x²)/2

= x + x/2 + y²/(2x)

= 3x/2 + y²/(2b²/y), by (1)

or, X = 3x/2 + y³/(2b²)

& Y = y + (1 + y₁²)/y₂

= y + (1 + y²/x²)/(2y/x²)

= y + x²/(2y) + y/2

= 3y/2 + x²/(2b²/x), by (1)

or, Y = 3y/2 + x³/(2b²)

Then X + Y

= 3/2 (x + y) + 1/(2b²) (x³ + y³)

= 1/(2b²) {3b² (x + y) + x³ + y³}

= 1/(2b²) {3xy (x + y) + x³ + y³}, by (1)

= 1/(2b²) (x³ + 3x²y + 3xy² + y³)

or, X + Y = 1/(2b²) * (x + y)³ ..... (3)

Similarly, we can obtain

X - Y = - 1/(2b²) * (x - y)³ ..... (4)

From (3) and (4), we get

(X + Y)^(2/3) - (X - Y)^(2/3)

= {(x + y)² - (x - y)²}/(2b²)^(2/3)

= 4xy/(a²)^(2/3), since b² = a²/2

= 2a²/(a²)^(2/3), since 2xy = a²

= 2 a^(2/3)

Therefore the evolute is

(x + y)^(2/3) - (x - y)^(2/3) = 2 a^(2/3)

Hence proved.

Note:

dy/dx = y₁

d²y/dx² = y₂