Differential Equations
Geometric Problems Leading to DEs
Grade 12
Question:
<p>A curve is such that the portion of the X-axis cut off between the origin and tangent at a point is twice the abscissa and which passes through the point <span>\((1, 2)\)</span>. The equation of the curve is</p>
<p>(a) <span>\(xy = 1\)</span></p>
<p>(b) <span>\(xy = 2\)</span></p>
<p>(c) <span>\(xy = 3\)</span></p>
<p>(d) <span>\(xy = 0\)</span></p>
Step-by-Step Solution
Key Concept: Translate the geometric condition about the tangent line into a differential equation by finding where the tangent meets the X-axis.
<p><strong>Step 1:</strong> Let the point on the curve be <span>$(x, y)$</span>. The equation of the tangent at this point is <span>$Y - y = \frac{dy}{dx}(X - x)$</span>.</p><p><strong>Step 2:</strong> The tangent meets the X-axis where <span>$Y = 0$</span>, giving <span>$X = x - \frac{y}{dy/dx}$</span>.</p><p><strong>Step 3:</strong> The portion of X-axis cut off between origin and this point is <span>$x - \frac{y}{dy/dx}$</span>, which equals <span>$2x$</span>.</p><p><strong>Step 4:</strong> Therefore: <span>$x - \frac{y}{dy/dx} = 2x$</span>, which gives <span>$\frac{dy}{dx} = -\frac{y}{x}$</span>.</p><p><strong>Step 5:</strong> Separating variables: <span>$\frac{dy}{y} = -\frac{dx}{x}$</span>, so <span>$\log y = -\log x + C$</span>, thus <span>$xy = k$</span>.</p><p><strong>Step 6:</strong> Using <span>$(1, 2)$</span>: <span>$1 \cdot 2 = k$</span>, so <span>$k = 2$</span>.</p><p>∴ Answer is B.</p>
Correct Answer: B