Question

The two curves x³ – 3xy² + 2 = 0 and 3x²y – y³ – 2 = 0 intersect at an angle of ______.

Options

• `pi/4`

• `pi/3`

• `pi/2`

• `pi/6`

Answer 1

The two curves x³ – 3xy² + 2 = 0 and 3x²y – y³ – 2 = 0 intersect at an angle of `pi/2`.

Explanation:

The given curves are x³ – 3xy² + 2 = 0  .....(i)

And 3x²y – y³ – 2 = 0   ......(ii)

Differentiating equation (i) w.r.t. x, we get

`3x^2 - 3 * (x * 2y  "dy"/"dx" + y^2 * 1)` = 0

⇒ `x^2 - 2xy "dy"?'dx" - y^2` = 0

⇒ `2xy "dy"/"dx"` = x² – y²

∴ `"dy"/"dx" = (x^2 - y^2)/(2xy)`

So slope of the curve m₁ = `(x^2 - y^2)/(2xy)`

Differentiating equation (ii) w.r.t. x, we get

`3[x^2 "dy"/"dx" + y * 2x] - 3y^2 * "dy"/"dx"` = 0

`x^2 "dy"/"dx" + 2xy - y^2 "dy"/"dx"` = 0

⇒ `(x^3 - y^2) "dy"/"dx"` = – 2xy

∴ `"dy"/"dx" = (-2xy)/(x^2 - y^2)`

So the slope of the curve m₂ = `(-2xy)/(x^2 - y^2)`

Now m₁ × m₂ = `(x^2 - y^2)/(2xy) xx (-2xy)/(x^2 - y^2)` = – 1

So the angle between the curves is `pi/2`.