Differential Equations
Geometric applications of differential equations
Grade 12
Question:
<p>The equation of normal of \(P(x, y)\) is<br>\((Y - y) = \dfrac{-1}{\dfrac{dy}{dx}}(X - x)\)</p><p>If the normal at every point of a curve passes through the origin \((0,0)\) and \(A\left(x + y\dfrac{dy}{dx}, 0\right)\) and \(B\left(0, y + x\dfrac{dy}{dx}\right)\) are points on the axes, which of the following are correct?</p>
<p>(a) The curve passes through origin</p>
<p>(b) The curve is a circle centered at origin</p>
<p>(c) \(A\) lies on the x-axis</p>
<p>(d) \(B\) lies on the y-axis</p>
Step-by-Step Solution
Key Concept: If the normal at every point passes through the origin, substitute (0,0) into the normal equation to get the differential equation y + x(dy/dx) = 0, then verify which statements about points A and B follow from this condition.
<p><strong>Step 1:</strong> Write the normal equation at P(x,y): $(Y - y) = \frac{-1}{dy/dx}(X - x)$</p><p><strong>Step 2:</strong> Since the normal passes through origin (0,0), substitute X=0, Y=0:</p><p>$(0 - y) = \frac{-1}{dy/dx}(0 - x)$</p><p>$-y = \frac{x}{dy/dx}$</p><p>$y\frac{dy}{dx} = -x$</p><p><strong>Step 3:</strong> This gives us: $x + y\frac{dy}{dx} = 0$ (Key differential equation)</p><p><strong>Step 4:</strong> Analyze point A: $A(x + y\frac{dy}{dx}, 0)$. Since $x + y\frac{dy}{dx} = 0$, point A is at origin (0, 0). ✓</p><p><strong>Step 5:</strong> For point B: $B(0, y + x\frac{dy}{dx})$. From $x + y\frac{dy}{dx} = 0$, we get $y\frac{dy}{dx} = -x$. This does NOT necessarily mean $y + x\frac{dy}{dx} = 0$. Point B is generally not at origin. ✗</p><p><strong>Step 6:</strong> The DE $y dy + x dx = 0$ integrates to $x^2 + y^2 = c$ (circles centered at origin). ✓</p><p>∴ Answers: <strong>a, d</strong> (Normal passes through origin; curve represents concentric circles)</p>
Correct Answer: a, d