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 (English)

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 $k$ .

00:00 / 00:00

Conceptually Similar Problems

MathematicsHomogeneous Differential EquationsJEE Advanced 1999Moderate

View in:English Hindi

A curve passing through the point (1, 1) has the property that the perpendicular distance of the origin from the normal at any point P of the curve is equal to the distance of P from the x-axis. Determine the equation of the curve.

Answer:

x^{2} + y^{2} - 2 x = 0

x = 1

MathematicsVariable Separable MethodJEE Advanced 2005Moderate

View in:English Hindi

If length of tangent at any point on the curve $y = f (x)$ intercepted between the point and the x-axis is of length 1. Find the equation of the curve.

Answer:

ln ∣ \frac{1 - 1 - y ^{2}}{y} ∣ + 1 - y^{2} = \pm x + c

MathematicsFormation of Differential EquationsJEE Main 2007Moderate

View in:English Hindi

The differential equation of all circles passing through the origin and having their centres on the $x$ -axis is

(A)

y^{2} = x^{2} + 2 x y \frac{d y}{d x}

(B)

y^{2} = x^{2} - 2 x y \frac{d y}{d x}

(C)

x^{2} = y^{2} + x y \frac{d y}{d x}

(D)

x^{2} = y^{2} + 3 x y \frac{d y}{d x}

Answer:A

MathematicsTangents, Normals and Rate MeasureJEE Main 2005Moderate

View in:English Hindi

The normal to the curve $x = a (cos θ + θ sin θ)$ , $y = a (sin θ - θ cos θ)$ at any point ' $θ$ ' is such that

(A)

it passes through the origin

(B)

it makes an angle

\frac{π}{2} + θ

with the

x

-axis

(C)

it passes through

(a \frac{π}{2}, - a)

(D)

it is at a constant distance from the origin

Answer:D

MathematicsVariable Separable MethodJEE Advanced 2006Moderate

View in:English Hindi

A curve $y = f (x)$ passes through (1, 1) and at $P (x, y)$ , tangent cuts the x-axis and y-axis at A and B respectively such that $B P : A P = 3 : 1$ , then

* Multiple Correct Options

(A)

equation of curve is

x y^{'} - 3 y = 0

(B)

normal at (1, 1) is

x + 3 y = 4

(C)

curve passes through (2, 1/8)

(D)

equation of curve is

x y^{'} + 3 y = 0

Answer:C, D

MathematicsEquation of Tangent and NormalJEE Advanced 1991Moderate

View in:English Hindi

Three normals are drawn from the point $(c, 0)$ to the curve $y^{2} = x$ . Show that $c$ must be greater than $1/2$ . One normal is always the x-axis. Find $c$ for which the other two normals are perpendicular to each other.

Answer:0.75

MathematicsTangents, Normals and Rate MeasureJEE Advanced 1993Moderate

View in:English Hindi

Find the equation of the normal to the curve $y = (1 + x)^{y} + sin^{- 1} (sin^{2} x)$ at $x = 0$

Answer:

x + y = 1

MathematicsVariable Separable MethodJEE Advanced 1996Moderate

View in:English Hindi

A curve $y = f (x)$ passes through the point $P (1, 1)$ . The normal to the curve at $P$ is $a (y - 1) + (x - 1) = 0$ . If the slope of the tangent at any point on the curve is proportional to the ordinate of the point, determine the equation of the curve. Also obtain the area bounded by the $y$ -axis, the curve and the normal to the curve at $P$ .

Answer:

y = e^{a (x - 1)}

; 1 sq. unit

MathematicsVariable Separable MethodJEE Advanced 1998Moderate

View in:English Hindi

A curve $C$ has the property that if the tangent drawn at any point $P$ on $C$ meets the co-ordinate axes at $A$ and $B$ , then $P$ is the mid-point of $A B$ . The curve passes through the point $(1, 1)$ . Determine the equation of the curve.

Answer:

x y = 1

MathematicsTangents, Normals and Rate MeasureJEE Advanced 1983Moderate

View in:English Hindi

The normal to the curve $x = a (cos θ + θ sin θ)$ , $y = a (sin θ - θ cos θ)$ at any point ' $θ$ ' is such that

(A)

it makes a constant angle with the

x

-axis

(B)

it passes through the origin

(C)

it is at a constant distance from the origin

(D)

none of these

Answer:C