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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`
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Solution
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`.
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