Question

If a curve passes through the point (1, 1) at any point (x, y) on the curve, the product of the slope of its tangent and x co-ordinate of the point is equal to the y co-ordinate of the point, then the curve also passes through the point

• A. (-1, 2)
• B. (2, 2)
• C. \((\sqrt3,0)\)
• D. (3, 0)

Hint:

When solving a differential equation of the form \( \frac{dy}{dx} = \frac{y}{x} \), separate the variables and integrate both sides. Using the initial condition, you can find the constant and determine the equation of the curve.

Answer 1

The Correct Option is B

The correct answer is: (B): (2, 2).

We are given that the curve passes through the point \( (1, 1) \) and that at any point \( (x, y) \) on the curve, the product of the slope of its tangent and the x-coordinate of the point is equal to the y-coordinate of the point. This can be written as:

\( \frac{dy}{dx} \cdot x = y \)

Step 1: Set up the differential equation

The given relationship between the slope of the tangent and the coordinates of the curve is:

\( \frac{dy}{dx} = \frac{y}{x} \)

Step 2: Solve the differential equation

We now solve this first-order differential equation. We can separate the variables:

\( \frac{dy}{y} = \frac{dx}{x} \)

Now integrate both sides:

\( \int \frac{dy}{y} = \int \frac{dx}{x} \)

The integrals are straightforward:

\( \ln |y| = \ln |x| + C \)

Step 3: Solve for \( y \)

Exponentiating both sides gives:

\( |y| = C|x| \)

Thus, the general solution is:

\( y = Cx \)

Step 4: Use the given point to find \( C \)

We are told that the curve passes through the point \( (1, 1) \). Substituting \( x = 1 \) and \( y = 1 \) into the equation \( y = Cx \) gives:

\( 1 = C \cdot 1 \)

Thus, \( C = 1 \), and the equation of the curve is:

\( y = x \)

Step 5: Check for another point on the curve

Now, substitute \( x = 2 \) into the equation \( y = x \) to get:

\( y = 2 \)

Conclusion:
The curve passes through the point \( (2, 2) \), so the correct answer is (B): (2, 2).