A normal is drawn at a point P(x, y) of...

MathematicsVariable Separable MethodJEE Advanced 1994Moderate

A normal is drawn at a point $P (x, y)$ of a curve. It meets the x-axis at Q. If PQ is of constant length $k$ , then show that the differential equation describing such curves is $y \frac{d y}{d x} = \pm k^{2} - y^{2}$ . Find the equation of such a curve passing through (0, k).

Answer:

x^{2} + y^{2} = k^{2}

Visualized Solution (Hindi)

Visualizing the Geometric Setup

Let the curve be $y = f (x)$ .
Point of interest: $P (x, y)$ .
Normal line at $P$ meets the x-axis at $Q$ .
Given condition: Length $P Q = k$ (constant).

Finding the Slope of the Normal

Slope of tangent at $P (x, y)$ is $m = \frac{d y}{d x}$ .
Since normal is perpendicular to tangent, slope of normal is $m_{n or ma l} = - \frac{1}{m} = - \frac{1}{d y / d x}$ .

Equation of the Normal Line

Equation of normal at $P (x, y)$ using point-slope form:
$Y - y = - \frac{1}{d y / d x} (X - x)$

Locating Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q on the X-axis

Point $Q$ lies on the x-axis, so set $Y = 0$ in the normal equation:
$0 - y = - \frac{1}{d y / d x} (X - x)$
$- y = - \frac{1}{d y / d x} (X - x)$

Solving for Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

Rearranging to find $X$ :
$y \frac{d y}{d x} = X - x \Rightarrow X = x + y \frac{d y}{d x}$
Coordinates of $Q$ are $(x + y \frac{d y}{d x}, 0)$ .

Applying the Constant Length Condition

Distance $P Q = (x_{Q} - x_{P})^{2} + (y_{Q} - y_{P})^{2} = k$
$(x + y \frac{d y}{d x} - x)^{2} + (0 - y)^{2} = k^{2}$
$(y \frac{d y}{d x})^{2} + y^{2} = k^{2}$

Deriving the Differential Equation

$(y \frac{d y}{d x})^{2} = k^{2} - y^{2}$
Taking square root on both sides:
$y \frac{d y}{d x} = \pm k^{2} - y^{2}$
This is the required differential equation.

Variable Separation for Integration

To find the curve, separate the variables:
$\frac{y}{k ^{2} - y ^{2}} d y = \pm d x$
Integrating both sides:
$\int \frac{y}{k ^{2} - y ^{2}} d y = \pm \int d x$

Solving the Integral

Substitution: Let $k^{2} - y^{2} = t^{2} \Rightarrow - 2 y d y = 2 t d t \Rightarrow y d y = - t d t$ .
LHS: $\int \frac{- t d t}{t} = - \int d t = - t = - k^{2} - y^{2}$ .
RHS: $\pm \int d x = \pm x + C$ .
General Solution: $- k^{2} - y^{2} = \pm x + C$ .

Applying the Initial Condition

Curve passes through $(0, k)$ . Substitute $x = 0, y = k$ :
$- k^{2} - k^{2} = \pm 0 + C \Rightarrow 0 = C$ .
So, the equation is $- k^{2} - y^{2} = \pm x$ .

Final Equation of the Curve

Squaring both sides:
$k^{2} - y^{2} = x^{2}$
Rearranging terms:
$x^{2} + y^{2} = k^{2}$
The curve is a circle with center $(0, 0)$ and radius $k$ .

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