Find the equation of a curve passing through the point $(1,\,1)$, if the tangent drawn at any point $P(x,\,y)$ on the curve meets the co-ordinate axes at $A$ and $B$ such that $P$ is the mid-point of $AB$.

Find the equation of a curve passing through the point (1,1), if the tangent drawn at any point P(x,y) on the

NCERT Exemplar · Q32

(1,\,1)

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Concept understanding — Separation Of Variables

Separation of Variables: From Intuition to Precision

Imagine you're baking a cake. The recipe says "mix the dry ingredients separately, then add the wet ones." You keep things that belong together together, and things that don't apart — until the right moment. Separation of Variables does exactly that for certain kinds of equations.

The Core Intuition

Some equations involve two different kinds of change happening at once. Think of a cup of hot coffee cooling down. The rate at which it cools depends on:

The temperature difference between the coffee and the room (a function of time)
The surface area of the cup (a function of shape, not time)

These two influences are tangled together in one equation. Separation of Variables is the mathematical trick that untangles them — it lets you handle the time part first, then the space part separately.

The Precise Statement

Separation of Variables applies to ordinary differential equations (ODEs) of the form:

\frac{dy}{dx} = f(x) \cdot g(y)

where the right-hand side is a product of a function of $x$ alone and a function of $y$ alone. The method works in three clean steps:

\frac{dy}{dx} = f(x) \cdot g(y) \quad \Longrightarrow \quad \frac{1}{g(y)} \, dy = f(x) \, dx

Step 1: Separate. Multiply both sides by $dx$ and divide by $g(y)$ (assuming $g(y) \neq 0$ ). This moves all $y$ 's to one side and all $x$ 's to the other.

Step 2: Integrate. Put an integral sign on both sides:

\int \frac{1}{g(y)} \, dy = \int f(x) \, dx

Step 3: Solve. Evaluate both integrals and solve for $y$ explicitly if possible.

Watch out

You cannot separate if the equation is not in product form. For example, $\frac{dy}{dx} = x + y$ cannot be separated — the sum $x+y$ is not a product $f(x)g(y)$ .

Why This Works

The justification is the chain rule in reverse. From $\frac{dy}{dx} = f(x)g(y)$ , rewrite it as:

\frac{1}{g(y)} \frac{dy}{dx} = f(x)

Now integrate both sides with respect to $x$ :

\int \frac{1}{g(y)} \frac{dy}{dx} \, dx = \int f(x) \, dx

The left side is a substitution waiting to happen: $\frac{dy}{dx} dx = dy$ , so you get $\int \frac{1}{g(y)} dy$ . That's the entire trick — the chain rule dressed up.

A Concrete Example

Solve $\frac{dy}{dx} = 2xy$ , with $y(0) = 3$ .

Step 1: Separate. Divide both sides by $y$ (assuming $y \neq 0$ ):

\frac{1}{y} \, dy = 2x \, dx

Step 2: Integrate.

\int \frac{1}{y} \, dy = \int 2x \, dx \quad\Longrightarrow\quad \log|y| = x^2 + C

Step 3: Solve for $y$ .

|y| = e^{x^2 + C} = e^C \cdot e^{x^2}

Let $A = \pm e^C$ (absorbing the absolute value): $y = A e^{x^2}$ . Now use $y(0) = 3$ : $3 = A e^{0} = A$ , so $A = 3$ .

Final answer: $\boxed{y = 3e^{x^2}}$ …

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The key idea is to translate the geometric condition “P is the midpoint of AB” into a differential equation using the tangent line equation. Solving that ODE by separating variables and applying the given point yields the rectangular hyperbola $xy = 1$ .

Why separation of variables works here

The problem gives a relationship between a curve and its tangent line. Every tangent line at a point $P(x,y)$ has a slope $\frac{dy}{dx}$ . The intercepts of that line on the axes can be expressed in terms of $x$ , $y$ , and $\frac{dy}{dx}$ . The condition that $P$ is the midpoint of those intercepts then becomes an equation linking $x$ , $y$ , and $\frac{dy}{dx}$ — a first-order differential equation.

That equation turns out to be separable: we can rearrange it so that all $y$ ‑terms (including $dy$ ) are on one side and all $x$ ‑terms (including $dx$ ) are on the other. Then we integrate both sides. The constant of integration is fixed by the given point $(1,1)$ .

Step‑by‑step solution

1. Write the tangent line equation

At a point $P(x,y)$ on the curve, the slope is $\frac{dy}{dx}$ . The equation of the tangent line in point‑slope form is:

Y - y = \frac{dy}{dx}\,(X - x)

where $(X,Y)$ are the coordinates of any point on the line.

2. Find the intercepts $A$ and $B$

$x$ ‑intercept (point $A$ ): set $Y = 0$ and solve for $X$ .

0 - y = \frac{dy}{dx}\,(X - x) \quad\Rightarrow\quad -y = \frac{dy}{dx}\,(X - x)

X - x = -\frac{y}{\frac{dy}{dx}} \quad\Rightarrow\quad X = x - \frac{y}{\frac{dy}{dx}}

So $A = \left(x - \dfrac{y}{\frac{dy}{dx}},\; 0\right)$ .

$y$ ‑intercept (point $B$ ): set $X = 0$ and solve for $Y$ .

Y - y = \frac{dy}{dx}\,(0 - x) \quad\Rightarrow\quad Y - y = -x\,\frac{dy}{dx}

Y = y - x\,\frac{dy}{dx}

So $B = \left(0,\; y - x\,\frac{dy}{dx}\right)$ .

Tip

A quick check: if the slope is negative (as it often is for curves passing through $(1,1)$ with this property), both intercepts are positive — which matches the geometry.

3. Apply the midpoint condition

$P(x,y)$ is the midpoint of $AB$ . The midpoint formula gives:

x = \frac{\left(x - \dfrac{y}{\frac{dy}{dx}}\right) + 0}{2}, \qquad y = \frac{0 + \left(y - x\,\frac{dy}{dx}\right)}{2}

Take the first equation:

x = \frac{1}{2}\left(x - \frac{y}{\frac{dy}{dx}}\right)

Multiply by 2:

2x = x - \frac{y}{\frac{dy}{dx}}

x = -\frac{y}{\frac{dy}{dx}}

\frac{dy}{dx} = -\frac{y}{x}

The second equation will give the same result — it’s consistent. …

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Q.Find the equation of a curve passing through the point $(1,\,1)$ , if the tangent drawn at any point $P(x,\,y)$ on the curve meets the co-ordinate axes at $A$ and $B$ such that $P$ is the mid-point of $AB$ .

Concept understanding — Separation Of Variables

Separation of Variables: From Intuition to Precision

The Core Intuition

The Precise Statement

Why This Works

A Concrete Example

Why separation of variables works here

Step‑by‑step solution

Method: Turning a tangent-geometry condition into a separable DE

Steps

Common Mistakes

Q.Find the equation of a curve passing through the point (1, 1)(1,\,1)(1,1), if the tangent drawn at any point P(x, y)P(x,\,y)P(x,y) on the curve meets the co-ordinate axes at AAA and BBB such that PPP is the mid-point of ABABAB.

Why separation of variables works here

Step‑by‑step solution

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Q.Find the equation of a curve passing through the point $(1,\,1)$ , if the tangent drawn at any point $P(x,\,y)$ on the curve meets the co-ordinate axes at $A$ and $B$ such that $P$ is the mid-point of $AB$ .

Unlock everything free for 14 days