Question:medium

A normal is drawn at a point $P(x, y)$ of a curve $y = f(x)$. The normal meets the $X$ axis at $Q$. $l(PQ) = k \cdot$ (k is a constant) Then equation of the curve through $(0, k)$ is

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A curve where the distance from a point to a fixed axis along the normal is constant is a circle.
Updated On: May 14, 2026
  • $x^2 + y^2 = k^2$
  • $(1+k)x^2 + y^2 = k^2$
  • $x^2 + (1+k^2)y^2 = k^2$
  • $x^2 + 2y^2 = 2k^2$
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The Correct Option is A

Solution and Explanation

Step 1: Understanding the Concept:
We are given a geometric property involving the normal to a curve. We need to formulate a differential equation based on this property and solve it to find the family of curves.
Step 2: Key Formula or Approach:
The equation of the normal at a point $P(x, y)$ to the curve $y = f(x)$ is $Y - y = -\frac{dx}{dy}(X - x)$.
Find the coordinates of $Q$ where the normal intersects the X-axis (set $Y=0$).
Use the distance formula for $l(PQ)$ and set it equal to $k$ to form a differential equation.
Step 3: Detailed Explanation:
The equation of the normal at $P(x, y)$ is:
\[ Y - y = -\frac{1}{y'} (X - x) = -\frac{dx}{dy} (X - x) \] It meets the X-axis at $Q(X_0, 0)$. Substitute $Y = 0$:
\[ -y = -\frac{dx}{dy} (X_0 - x) \] \[ y \frac{dy}{dx} = X_0 - x \implies X_0 = x + y \frac{dy}{dx} \] So, the coordinates of $Q$ are $(x + y \frac{dy}{dx}, 0)$.
The distance $l(PQ)$ is given by the distance formula:
\[ l(PQ) = \sqrt{(x + y \frac{dy}{dx} - x)^2 + (0 - y)^2} \] \[ l(PQ) = \sqrt{\left(y \frac{dy}{dx}\right)^2 + y^2} = |y| \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \] We are given that $l(PQ) = k$. Squaring both sides:
\[ y^2 \left[ 1 + \left(\frac{dy}{dx}\right)^2 \right] = k^2 \] \[ y^2 + y^2 \left(\frac{dy}{dx}\right)^2 = k^2 \] \[ y^2 \left(\frac{dy}{dx}\right)^2 = k^2 - y^2 \] Taking the square root:
\[ y \frac{dy}{dx} = \pm \sqrt{k^2 - y^2} \] Separate variables:
\[ \frac{y}{\sqrt{k^2 - y^2}} dy = \pm dx \] Integrate both sides:
\[ \int \frac{y}{\sqrt{k^2 - y^2}} dy = \int \pm 1 dx \] Let $t = k^2 - y^2$, then $dt = -2y dy$, so $y dy = -\frac{dt}{2}$.
\[ \int -\frac{1}{2\sqrt{t}} dt = \pm x + C \] \[ -\frac{1}{2} (2\sqrt{t}) = \pm x + C \] \[ -\sqrt{k^2 - y^2} = \pm x + C \] We are given that the curve passes through the point $(0, k)$. Substitute $x=0, y=k$:
\[ -\sqrt{k^2 - k^2} = \pm 0 + C \implies 0 = C \] So the equation simplifies to:
\[ -\sqrt{k^2 - y^2} = \pm x \] Squaring both sides to eliminate the sign and the radical:
\[ k^2 - y^2 = x^2 \] \[ x^2 + y^2 = k^2 \] Step 4: Final Answer:
The equation of the curve is $x^2 + y^2 = k^2$.
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