Differential Equations
Formation and Solution of Differential Equations
Grade 12
Question:
<p>The curve for which the intercept cut off by any tangent on Y-axis is proportional to the square of the ordinate of the point of tangency is</p>
<p>(a) \(\frac{C_1}{x} - \frac{C_2}{y} = 1\)</p>
<p>(b) \(\frac{C_1}{x} + \frac{C_2}{y} = 1\)</p>
<p>(c) \(\frac{C_1}{x} + \frac{C_2}{y} = \frac{1}{xy}\)</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Find the Y-intercept of the tangent line at a general point on the curve, apply the given proportionality condition, and solve the resulting differential equation.
<p><strong>Step 1:</strong> Let <i>P(x, y)</i> be any point on the curve.</p><p><strong>Step 2:</strong> The equation of the tangent at <i>P(x, y)</i> is:</p><p>$$Y - y = \frac{dy}{dx}(X - x)$$</p><p><strong>Step 3:</strong> To find the Y-intercept, set <i>X = 0</i>:</p><p>$$Y - y = \frac{dy}{dx}(0 - x)$$</p><p>$$Y = y - x\frac{dy}{dx}$$</p><p><strong>Step 4:</strong> The intercept on Y-axis (length from origin) is:</p><p>$$\text{Intercept} = y - x\frac{dy}{dx}$$</p><p><strong>Step 5:</strong> According to the problem, this is proportional to $y^2$:</p><p>$$y - x\frac{dy}{dx} \propto y^2$$</p><p>$$y - x\frac{dy}{dx} = ky^2$$</p><p><strong>Step 6:</strong> Solving this differential equation gives:</p><p>$$\frac{C_1}{x} + \frac{C_2}{y} = 1$$</p><p>∴ Answer is (b).</p>
Correct Answer: B