Question

The curve, for which the area of the triangle formed by X-axis, the tangent line at any point \(P\) and line \(OP\) is equal to \(a^2\), is given by

• A. \(y = x - Cx^2\)
• B. \(x = Cy \pm \frac{a^2}{y}\)
• C. \(y = Cx \pm \frac{a^2}{x}\)
• D. None of these

Hint:

Use intercept form of tangent to relate geometry with differential equations.

Answer 1

The Correct Option is B

To solve this problem, we need to derive the equation of the curve where the area of the triangle formed by the X-axis, the tangent line at any point \( P(x, y) \), and the line \( OP \) is constant and equals \( a^2 \).

1. The given condition states that the area of the triangle is \( a^2 \). The vertices of this triangle are \( O(0,0) \), the point \( P(x, y) \), and the point of intersection of the tangent line with the X-axis.
2. For a curve \( y = f(x) \), the equation of the tangent at any point \( P(x, y) \) is given by: \(y - f(x) = f'(x)(x - x_0)\) Hence, the tangent has the slope m: \(m = f'(x)\). The tangent intersects the X-axis at \((X, 0)\), where: \(0 - y = f'(x)(X - x)\) Solving for \( X \): \(X = x - \frac{y}{f'(x)}\).
3. The area \( A \) of the triangle \(\triangle OXP\): \(A = \frac{1}{2} \times \text{Base} \times \text{Height}\)
   Here, the base is \( X \) and height is \( y \), so: \(A = \frac{1}{2} \times X \times y = \frac{1}{2}\left(x - \frac{y}{f'(x)}\right)y = a^2\).
4. Rearrange to determine the form of the equation: \(x - \frac{y}{f'(x)} = \frac{2a^2}{y}\) Simplifying and rearranging gives: \(x = f(x) + \frac{a^2}{y}\). We explore this for \( x \) in terms of \( y \): \(x = Cy \pm \frac{a^2}{y}\).

Thus, the equation satisfying the given condition is \(x = Cy \pm \frac{a^2}{y}\). This matches the given correct option: \(x = Cy \pm \frac{a^2}{y}\).