Applications of Derivatives
Tangent and Normal
Grade 12
Question:
<p>The equation of the curve in which the perpendicular from the origin upon the tangent is equal to the abscissa of the point of contact is</p>
<p>(A) \(x^2 + y^2 = cx\)</p>
<p>(B) \(x^2 + y^2 = cy\)</p>
<p>(C) \(x^2 + y^2 = c\)</p>
<p>(D) None of these</p>
Step-by-Step Solution
Key Concept: Set up the perpendicular distance formula from origin to tangent line and equate it to the x-coordinate of the point of contact.
<p>Let the point of contact be $(x_0, y_0)$ on the curve. The tangent at this point has the property that the perpendicular distance from the origin equals the abscissa $x_0$.</p><p>For a curve $y = f(x)$, the equation of tangent at $(x_0, y_0)$ is: $Y - y_0 = \frac{dy}{dx}(X - x_0)$</p><p>Perpendicular distance from origin to tangent: $d = \frac{|y_0 - x_0\frac{dy}{dx}|}{\sqrt{1 + (\frac{dy}{dx})^2}} = x_0$</p><p>This leads to the differential equation which solves to $x^2 + y^2 = cy$.</p>
Correct Answer: B