Question

The curve passing through the point (1, 2) given that the slope of the tangent at any point (x, y) is \(\frac{2x}{y}\) represents

• A. Circle
• B. Parabola
• C. Ellipse
• D. Hyperbola

Answer 1

The Correct Option is D

The curve passing through the point (1, 2) given that the slope of the tangent at any point (x, y) is \(\frac{2x}{y}\) represents which type of curve?

The slope of the tangent at any point is \(\frac{dy}{dx} = \frac{2x}{y}\).

We can separate variables to solve this differential equation:

\(y \, dy = 2x \, dx\)

Integrate both sides:

\(\int y \, dy = \int 2x \, dx\)

\(\frac{y^2}{2} = x^2 + C\)

\(y^2 = 2x^2 + 2C\)

Since the curve passes through (1, 2), we can substitute x = 1 and y = 2 to find C:

\(2^2 = 2(1)^2 + 2C\)

\(4 = 2 + 2C\)

\(2 = 2C\)

\(C = 1\)

So the equation of the curve is \(y^2 = 2x^2 + 2\). Rearranging, we get \(y^2 - 2x^2 = 2\), or \(\frac{y^2}{2} - x^2 = 1\).

This is the equation of a hyperbola.

Therefore, the correct option is (D) Hyperbola.

Answer 2

Integrating both sides:

$$ \int y \, dy = \int 2x \, dx \implies \frac{y^2}{2} = x^2 + c \implies y^2 = 2x^2 + 2c \implies y^2 - 2x^2 = k, $$

where $ k = 2c $. Since the curve passes through $ (1, 2) $, substitute $ x = 1 $ and $ y = 2 $:

$$ 2^2 - 2(1^2) = 4 - 2 = 2 = k. $$

Thus, the equation of the curve is:

$$ y^2 - 2x^2 = 2. $$

This equation is of the form $ y^2 - ax^2 = b $, which represents a hyperbola. 

The final answer is $ {\text{Hyperbola}} $.