A curve passes through (1, 1) such that the triangle formed by the coordinate axes and the tangent at any point of the curve is in the first quadrant and has its area equal to 2. What will be the equation of the curve?a)xy = 2b)xy = -1c)x – y = 2d)x + y = 2Correct answer is option 'D'. Can you explain this answer?

Question

A curve passes through (1, 1) such that the triangle formed by the coordinate axes and the tangent at any point of the curve is in the first quadrant and has its area equal to 2. What will be the equation of the curve?a)xy = 2b)xy = -1c)x &ndash; y = 2d)x + y = 2Correct answer is option 'D'. Can you explain this answer?

Answer

To find the equation of the curve, we need to determine the slope of the tangent line at any point on the curve.

Let's assume the equation of the curve is y = f(x).

At the point (1, 1), the slope of the tangent line can be determined as follows:

slope = f'(1) = tan(θ)

where θ is the angle between the positive x-axis and the tangent line.

Since the triangle formed by the coordinate axes and the tangent line is in the first quadrant, tan(θ) > 0.

Now, let's consider a small change in x, denoted as Δx. The corresponding change in y can be expressed as Δy = f'(1) * Δx.

The area of the triangle formed by the coordinate axes and the tangent line can be calculated as:

Area = (1/2) * Δx * Δy
= (1/2) * Δx * f'(1) * Δx
= (1/2) * (f'(1)) * (Δx)^2

Given that the area is equal to 2, we have:

2 = (1/2) * (f'(1)) * (Δx)^2

Simplifying, we get:

4 = (f'(1)) * (Δx)^2

Since Δx can be any small value, the only way the above equation holds true is if f'(1) = 4.

Now, let's integrate f'(x) = 4 to find the equation of the curve:

∫ f'(x) dx = ∫ 4 dx
f(x) = 4x + C

where C is the constant of integration.

Using the point (1, 1) on the curve, we can solve for C:

1 = 4(1) + C
C = -3

Therefore, the equation of the curve is:

f(x) = 4x - 3

Multiplying both sides by x, we get:

xy = 4x^2 - 3x

So, the correct option is:

b) xy = -1

Answer

The equation of tangent to the curve y = f(x), at point (x, y), is
Y – y = dy/dx * (X – x) …..(1)
Where it meets the x axis, Y = 0 and X = (x – y/(dy/dx))
Where it meets the y axis, X = 0 and Y = (y – x/(dy/dx))
Also, the area of the triangle formed by (1) with the coordinate axes is 2, so that,
(x – y/(dy/dx))* (y – x/(dy/dx)) = 4
Or, (y – x/(dy/dx))² – 4dy/dx = 0
Or, x²(dy/dx)² – 2(xy – 2)dy/dx + y² = 0
Solving for dy/dx we get,
dy/dx = [(xy – 2) ± √(1 – xy)]/ x²
Let, 1 – xy = t²
⇒ x(dy/dx) + y = -2t(dt/dx)
⇒ x²(dy/dx) = t² – 1 – 2tx(dt/dx), so that (3) gives
t(x(dt/dx) – (t ± 1)) = 0
Hence, either t = 0
⇒ xy = 1 which is satisfied by (1, 1)
Or, x dt/dx = t ± 1
⇒ dx/x = dt/t ± 1
⇒ t ± 1 = cx
For x = 1, y = 1 and t = 0
⇒ c = ± 1, so the solution is
t = ± (x – 1) => t² = (x – 1)²
Or, 1 – xy = x² – 2x + 1
Or, x + y = 2
Thus, the two curves that satisfies are xy = 1 and x + y = 2

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