1.

What will be the value of angle between the curves x^2 – y^2 = 2a^2 and xv + y^2 = 4a^2?(a) π/2(b) π/4(c) π/6(d) π/3The question was asked in an internship interview.Question is taken from Calculus Application in portion Application of Calculus of Mathematics – Class 12

Answer»

Right option is (d) π/3

Easy explanation: X^2 – y^2 = 2a^2 ……….(1) and x^2 + y^2 = 4a^2 ……….(2)

Adding (1) and (2) we get, 2x^2 = 6a^2

Again, (2) – (1) gives,

2y^2 = 2a^2

Therefore, 2x^2 * 2y^2 = 6a^2 * 2a^2

4x^2y^2 = 12a^2

Or x^2y^2 = 3a^4

Or 2xy = ±2√3

Differentiating both side of (1) and (2) with RESPECT to x we get,

2x – 2y(dy/dx) = 0

Or dy/dx = x/y

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

Ordy/dx = -x/y

Let (x, y) be the point of intersection of the curves(1) and (2) and m1 and m2 be the slopes of the tangents to the curves (1) and (2) RESPECTIVELY at the point (x, y); then,

m1 = x/y and m2 = -x/y

Now the ANGLE between the curves (1) and (2) means the angle between the tangents to the CURVE at their point of intersection.

Therefore, if θ is the required angle between the curves (1) and (2), then

tanθ = |(m1 – m2)/(1 + m1m2)|

Putting the value of m1, m2 in the above equation we get,

tanθ = |2xy/(y^2 – x^2)|

As, 2xy = ±2√3a^2 and x^2 – y^2 = 2a^2

tanθ = |±2√3a^2/-2a^2|

Or tanθ = √3

Thus, θ = π/3.


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