Question:
The curve for which the length of the normal is equal to the length of the radius vector, are
The curve for which the length of the normal is equal to the length of the radius vector, are
Show Hint
Set up the differential equation and solve for the family of curves.
Updated On: Apr 23, 2026
- only circles
- only rectangular hyperbolas
- either circles or rectangular hyperbolas
- None of the above
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Formula / Definition}
\[ \text{Normal length} = y\sqrt{1 + \left(\frac{dy}{dx}\right)^2}, \text{Radius vector} = \sqrt{x^2 + y^2} \]
Step 2: Calculation / Simplification}
\(y^2\left[1 + \left(\frac{dy}{dx}\right)^2\right] = x^2 + y^2\)
\(y^2\left(\frac{dy}{dx}\right)^2 = x^2 \Rightarrow \frac{dy}{dx} = \pm \frac{x}{y}\)
\(y dy = \pm x dx \Rightarrow \frac{y^2}{2} = \pm \frac{x^2}{2} + c\)
\(x^2 + y^2 = c_1\) (circles) or \(x^2 - y^2 = c_2\) (rectangular hyperbolas)
Step 3: Final Answer
\[ \text{either circles or rectangular hyperbolas} \]
\[ \text{Normal length} = y\sqrt{1 + \left(\frac{dy}{dx}\right)^2}, \text{Radius vector} = \sqrt{x^2 + y^2} \]
Step 2: Calculation / Simplification}
\(y^2\left[1 + \left(\frac{dy}{dx}\right)^2\right] = x^2 + y^2\)
\(y^2\left(\frac{dy}{dx}\right)^2 = x^2 \Rightarrow \frac{dy}{dx} = \pm \frac{x}{y}\)
\(y dy = \pm x dx \Rightarrow \frac{y^2}{2} = \pm \frac{x^2}{2} + c\)
\(x^2 + y^2 = c_1\) (circles) or \(x^2 - y^2 = c_2\) (rectangular hyperbolas)
Step 3: Final Answer
\[ \text{either circles or rectangular hyperbolas} \]
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