JEE Advanced 1998

LEVELJEE Advanced

Animated Solution for Mathematics - Differential Equations: 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.

Progressive Hints

Hint 1

Visualize the geometry. You have a curve, a tangent, and two intercepts on the axes. The condition that

P

is the midpoint of the segment

A B

is a powerful geometric constraint. Can you draw this? The triangle formed by the origin and the intercepts is key to understanding the relationship between the coordinates.

Hint 2

Recall the point-slope form of a line:

Y - y = y^{'} (X - x)

. Use this to find the coordinates of

A

(where

Y = 0

) and

B

(where

X = 0

). Once you have these, the midpoint formula

x = \frac{x _{A} + x _{B}}{2}

and

y = \frac{y _{A} + y _{B}}{2}

will bridge the gap to your differential equation.

Hint 3

Start by setting up the midpoint equation for the

x

-coordinate:

x = \frac{( x - \frac{y}{y ^{'}} ) + 0}{2}

. Simplify this algebraic expression to isolate

y^{'}

, which is

\frac{d y}{d x}

. This will give you a separable differential equation that is the gateway to the final solution.

Visualized Solution

Visualizing the Problem

Let the curve be $C$ and a general point on it be $P (x, y)$ .
The tangent at $P$ meets the $X$ -axis at $A$ and the $Y$ -axis at $B$ .
Given: $P$ is the midpoint of the segment $A B$ .
The curve passes through the point $(1, 1)$ .

Equation of the Tangent at $P (x, y)$

The slope of the tangent at $P (x, y)$ is $m = \frac{d y}{d x} = y^{'}$ .
Equation of the tangent at $P (x, y)$ in point-slope form:
$Y - y = y^{'} (X - x)$

Finding the $X$ -intercept $A$

For point $A$ ( $X$ -intercept), set $Y = 0$ :
$- y = y^{'} (X - x)$
$X - x = - \frac{y}{y ^{'}}$
$X = x - \frac{y}{y ^{'}}$
So, $A = (x - \frac{y}{y ^{'}}, 0)$

Finding the $Y$ -intercept $B$

For point $B$ ( $Y$ -intercept), set $X = 0$ :
$Y - y = y^{'} (0 - x)$
$Y = y - x y^{'}$
So, $B = (0, y - x y^{'})$

Applying the Midpoint Condition

Since $P (x, y)$ is the midpoint of $A B$ , we equate the $x$ -coordinates:
$x = \frac{X _{A} + X _{B}}{2}$
$x = \frac{( x - \frac{y}{y ^{'}} ) + 0}{2}$

Forming the Differential Equation

Multiply by $2$ : $2 x = x - \frac{y}{y ^{'}}$
Subtract $x$ : $x = - \frac{y}{y ^{'}}$
Rearranging for $y^{'}$ : $y^{'} = - \frac{y}{x}$
Therefore, $\frac{d y}{d x} = - \frac{y}{x}$

Solving by Variable Separation

Separate the variables $x$ and $y$ :
Bring $y$ terms to one side and $x$ terms to the other.
$\frac{1}{y} d y = - \frac{1}{x} d x$

Integrating Both Sides

Integrate both sides:
$\int \frac{1}{y} d y = - \int \frac{1}{x} d x$
$ln ∣ y ∣ = - ln ∣ x ∣ + ln ∣ c ∣$
$ln ∣ y ∣ + ln ∣ x ∣ = ln ∣ c ∣ \Rightarrow ln ∣ x y ∣ = ln ∣ c ∣$
$x y = c$

Applying the Initial Condition

The curve passes through $(1, 1)$ .
Substitute $x = 1, y = 1$ into $x y = c$ :
$(1) (1) = c \Rightarrow c = 1$
The equation of the curve is $x y = 1$ .

Conclusion and Key Takeaway

Key Takeaway: The geometric property where the tangent midpoint lies on the curve leads to the differential equation $\frac{d y}{d x} = - \frac{y}{x}$ .
Final Equation: $x y = 1$ , which represents a rectangular hyperbola.

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The Sigma Insight: Variable Separable Method

Solution Diagram

Analyzing the Geometric Constraint

Imagine you are standing on a path defined by a function

y = f (x)

. At any point

P (x, y)

on this path, you draw a tangent line. This line acts as a mirror, reflecting the local behavior of the curve.

The tangent line strikes the

X

-axis at point

A

and the

Y

-axis at point

B

. The problem states that

P

is the midpoint of the segment

A B

. This geometric constraint implies that the curve is perfectly balanced between its intercepts.

Translating to Calculus

The equation of the tangent line at point

P (x, y)

with slope

y^{'} = \frac{d y}{d x}

is given by:

Y - y = y^{'} (X - x)

To find the intercepts, we set

Y = 0

to find

A

and

X = 0

to find

B

. This yields the coordinates:

A = (x - \frac{y}{y ^{'}}, 0), B = (0, y - x y^{'})

The Master Equation

We apply the midpoint condition, which states that the

x

-coordinate of

P

must be the average of the

x

-coordinates of

A

and

B

. This gives us the relation:

x = \frac{( x - \frac{y}{y ^{'}} ) + 0}{2}

Solving this equation, we obtain

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

, which simplifies to:

x = - \frac{y}{y ^{'}}

Rearranging this, we arrive at the fundamental differential equation:

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

Solving the Differential Equation

This equation indicates that the rate of change of

y

with respect to

x

is inversely proportional to

x

. We separate the variables to integrate:

\int \frac{1}{y} d y = - \int \frac{1}{x} d x

Integrating both sides, we obtain

ln ∣ y ∣ = - ln ∣ x ∣ + ln ∣ c ∣

, which simplifies to

ln ∣ x y ∣ = ln ∣ c ∣

, or:

x y = c

Final Calculation

Given that the curve passes through the point

(1, 1)

, we substitute these values to find the constant:

(1) (1) = c \Rightarrow c = 1

The resulting curve is $x y = 1$ , which is a rectangular hyperbola. Whenever you encounter a problem involving tangents and midpoints, you are likely looking at a hyperbola in disguise.

Common Traps & FAQs

Why is the derivative negative? Shouldn't the slope be positive?

That is a fantastic observation! The negative sign arises naturally from the geometry. If the tangent cuts the axes in the first quadrant, the line must have a negative slope to connect the positive

X

and

Y

intercepts. Trust the algebra—it captures the physical reality perfectly. The curve is decreasing as

x

increases, which is exactly what a negative derivative tells us.

Can I use the intercept form of a line

\frac{X}{a} + \frac{Y}{b} = 1

Absolutely! That is a brilliant shortcut. If you use the intercept form, you can relate

a

and

b

to the coordinates of

P (x, y)

using the midpoint condition

a = 2 x

and

b = 2 y

. Substituting these into the line equation

\frac{X}{2 x} + \frac{Y}{2 y} = 1

and ensuring the point

(x, y)

lies on it leads to the same result. It is often faster and shows deep conceptual mastery.

Why do we use

ln ∣ c ∣

instead of just

C

It is a matter of mathematical elegance and convenience. When you have

ln ∣ y ∣ = - ln ∣ x ∣ + C

, writing the constant as

ln ∣ c ∣

allows you to combine the logarithms using the property

ln (a) + ln (b) = ln (ab)

. It makes the final step of exponentiating both sides to remove the logarithms much cleaner and avoids messy algebraic manipulation.

Visualized Solution

Visualizing the Problem

Equation of the Tangent at $P (x, y)$

Finding the $X$ -intercept $A$

Finding the $Y$ -intercept $B$

Applying the Midpoint Condition

Forming the Differential Equation

Solving by Variable Separation

Integrating Both Sides

Applying the Initial Condition

Conclusion and Key Takeaway

The Sigma Insight: Variable Separable Method

Analyzing the Geometric Constraint

Translating to Calculus

The Master Equation

Solving the Differential Equation

Final Calculation

Similar Questions

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Visualized Solution

Visualizing the Problem

Equation of the Tangent at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(x,y)

Finding the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X-intercept .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Finding the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Y-intercept .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Applying the Midpoint Condition

Forming the Differential Equation

Solving by Variable Separation

Integrating Both Sides

Applying the Initial Condition

Conclusion and Key Takeaway

The Sigma Insight: Variable Separable Method

Analyzing the Geometric Constraint

Translating to Calculus

The Master Equation

Solving the Differential Equation

Final Calculation

Similar Questions

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Equation of the Tangent at $P (x, y)$

Finding the $X$ -intercept $A$

Finding the $Y$ -intercept $B$

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.

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

The normal to a curve at $P (x, y)$ meets the x-axis at $G$ . If the distance of $G$ from the origin is twice the abscissa of $P$ , then the curve is a

The slope of normal at any point $(x, y), x > 0, y > 0$ on the curve $y = y (x)$ is given by $\frac{x ^{2}}{x y - x ^{2} y ^{2} - 1}$ . If the curve passes through the point $(1, 1)$ , then $e \cdot y (e)$ is equal to

If a curve $y = f (x)$ passes through the point $(1, - 1)$ and satisfies the differential equation, $y (1 + x y) d x = x d y$ , then $f (- \frac{1}{2})$ is equal to :