Question

4) Find the tangent at origin of the given. curve y2 (2a-x) = x3

Answer 1

Answer:

The curve is symmetrical about the x-axis.

{only even powers of 'y' occur in the equation}

2. Origin : The curve passes through the origin.

{ there is no constant term in its equation.}

The tangents at the origin are y = 0,

{y = 0 equating to zero the lowest degree terms}

Origin is a cusp.

3. Asymptotes : The curve has an asymptote x = 2a.

{Co-eff. of y3 is absent, Co-eff. of y2 is an asymptote.}

4. Points : (a) curve meets the axes at (0,0) only.

(b) y2 = x3/(2a-x)

When x is - ve, y2 is - ve (y is imaginary) so that no portion of the curve lies to the left of the y-axis. Also when x> 2a, y2 is again-ve, so that no portion of the curve lies to the right of the line 3x = 2a.