Tangent at any point to a curve in the first quadrant meets the coordinate axes at A and B such that area of triangle OAB is always 2 square units. If the curve passes through (1, 1), thena)one such curve is x + y = 2b)one such curve is xy = 1c)one such curve is y = 1/(1+x2)d)the curve cannot be a straight lineCorrect answer is option 'A,B'. Can you explain this answer?

Question

Answer

The line given in option (1) satisfies the condition of the given problem and can thus be treated as a tangent to itself at any point.

⇒ (1) is true and (4) is false.

For curve in option (2):

At x = 1, tangent at (x₀, y₀)

Now,

x intercept = x₀ + x₀²y₀ = 2x₀

Thus, curves given in options (1) and (2) satisfy the condition of the problem.

Curve given in option (3) does not pass through (1, 1).

⇒ (3) is false.

Answer

To find the equation of the curve passing through (1,1) such that the area of triangle OAB is always 2 square units, let's analyze the given options:

a) one such curve is x * y = 2
b) one such curve is xy = 1
c) one such curve is y = 1/(1 + x^2)
d) the curve cannot be a straight line

Let's evaluate each option to determine if it satisfies the given condition.

a) x * y = 2
To find the equation of the tangent at any point (x, y) on this curve, we need to find the derivative dy/dx.

Taking the derivative of both sides with respect to x, we get:
y + x * dy/dx = 0
dy/dx = -y/x

The slope of the tangent at any point (x, y) on the curve is -y/x.

Now, let A and B be the x and y-intercepts of the tangent. The area of triangle OAB is given by:
Area = 1/2 * |A * B|

Since A and B are the x and y-intercepts, respectively, we have A = 1/y and B = 1/x.

Substituting these values, we get:
Area = 1/2 * |(1/y) * (1/x)|
= 1/2 * |1/(x * y)|
= 1/2 * |1/2|
= 1 square unit

Therefore, the area of the triangle is not always 2 square units for the curve x * y = 2. Hence, option 'a' is incorrect.

b) xy = 1
Similarly, taking the derivative dy/dx, we get:
y + x * dy/dx = 0
dy/dx = -y/x

The slope of the tangent at any point (x, y) on the curve is -y/x.

Using the same approach as above, we find that the area of triangle OAB is always 2 square units for this curve. Therefore, option 'b' is correct.

c) y = 1/(1 + x^2)
Taking the derivative dy/dx, we get:
dy/dx = -2x/(1 + x^2)^2

The slope of the tangent at any point (x, y) on the curve is -2x/(1 + x^2)^2.

Using the same approach as above, we find that the area of triangle OAB is not always 2 square units for this curve. Therefore, option 'c' is incorrect.

d) the curve cannot be a straight line
This statement is true because the area of triangle OAB will always be zero if the curve is a straight line passing through the origin (0,0). Therefore, option 'd' is correct.

In conclusion, the correct options are 'A' and 'D'. The curve can be represented by the equation xy = 1 or it cannot be a straight line.

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