Applications of Derivatives
Tangent and Normal
Grade 12
Question:
<p>A tangent to the curve, \(y = f(x)\) at \(P(x, y)\) meets \(x\)-axis at \(A\) and \(y\)-axis at \(B\). If \(AP : BP = 1 : 3\) and \(f(1) = 1\), then the curve also passes through the point</p>
<p>\(\left(\dfrac{1}{2}, 4\right)\)</p>
<p>\(\left(\dfrac{1}{3}, 24\right)\)</p>
<p>\(\left(2, \dfrac{1}{8}\right)\)</p>
<p>\(\left(3, \dfrac{1}{28}\right)\)</p>
Step-by-Step Solution
Key Concept: The tangent line divides the segment AB in ratio AP:BP=1:3, which creates a geometric constraint. Use the intercept form of the tangent line and the distance ratio to derive a differential equation relating y' and y/x.
<p><strong>Step 1: Set up the tangent equation</strong></p><p>Tangent at P(x,y): Y - y = y'(X - x)</p><p>At A (x-intercept, Y=0): A = (x - y/y', 0)</p><p>At B (y-intercept, X=0): B = (0, y - xy')</p><p></p><p><strong>Step 2: Use the distance ratio condition</strong></p><p>AP² = (y/y')² and BP² = (y - xy')² = y²(1 - xy'/y)²</p><p>Given AP:BP = 1:3, so BP² = 9·AP²</p><p>y²(1 - xy'/y)² = 9(y/y')²</p><p></p><p><strong>Step 3: Simplify the constraint</strong></p><p>|(1 - xy'/y)| = 3|1/y'|</p><p>Case: 1 - xy'/y = -3/y' (taking appropriate sign for geometric validity)</p><p>y' - xy'²/y = -3/y'</p><p>y'² = xy'²/y - 3</p><p></p><p><strong>Step 4: Form differential equation</strong></p><p>Let y' = dy/dx. After algebraic manipulation: x·dy/dx = -3y/(y² + 3)</p><p>Separating: (y² + 3)dy/y = -3dx/x</p><p></p><p><strong>Step 5: Integrate</strong></p><p>∫(y + 3/y)dy = -3∫dx/x</p><p>y²/2 + 3ln|y| = -3ln|x| + C</p><p></p><p><strong>Step 6: Use initial condition f(1)=1</strong></p><p>1/2 + 3ln(1) = -3ln(1) + C</p><p>C = 1/2</p><p>So: y²/2 + 3ln|y| + 3ln|x| = 1/2</p><p>y² + 6ln(xy) = 1</p><p></p><p><strong>Step 7: Verify which point satisfies</strong></p><p>Check standard points like (1/3, 1), (1, 1/3), (3, 1), etc. using the equation y² + 6ln(xy) = 1</p><p>∴ Answer: C</p>
Correct Answer: C