A curve $y=f(x)$ passes through the point $P(1,1)$. The equation of the normal at $P(1,1)$ to the curve $y=f(x)$ is $(x-1)+(y-1)=0$ and the slope of the tangent at any point on the curve is proportional to the ordinate of the point, then the equation of the curve is :

Question

Accepted Answer

To determine why Option 4 is the correct answer, we need to analyze the given conditions and how they relate to the equation of the curve.

1. **Understanding the Normal Equation:**
   The normal to the curve at point $ P(1,1) $ is given by the equation $(x-1) + (y-1) = 0$, which simplifies to $x + y = 2$. The slope of this normal line is $-1$.

2. **Slope of the Tangent:**
   The slope of the tangent at any point on the curve is proportional to the ordinate (the $y$-coordinate) of the point. This means that if the slope of the tangent is $m$, then $m = ky$ for some constant $k$.

3. **Finding the Slope at $P(1,1)$:**
   At point $P(1,1)$, the slope of the tangent is the negative reciprocal of the slope of the normal, which is $1$. Therefore, $ky = 1$ when $y = 1$, giving $k = 1$.

4. **Differential Equation:**
   The condition that the slope of the tangent is proportional to the ordinate can be expressed as $\frac{dy}{dx} = y$.

5. **Solving the Differential Equation:**
   The differential equation $\frac{dy}{dx} = y$ is a standard separable differential equation. Solving it, we get:
   $$
   \frac{dy}{y} = dx
   $$
   Integrating both sides, we have:
   $$
   \ln |y| = x + C
   $$
   Exponentiating both sides, we get:
   $$
   y = e^{x+C} = e^C \cdot e^x
   $$
   Let $e^C = A$, so $y = A e^x$.

6. **Applying Initial Condition:**
   The curve passes through $P(1,1)$, so substituting $x = 1$ and $y = 1$ into the equation $y = A e^x$, we get:
   $$
   1 = A e^1 \implies A = \frac{1}{e}
   $$
   Thus, the equation becomes:
   $$
   y = \frac{1}{e} e^x = e^{x-1}
   $$

7. **Conclusion:**
   The equation of the curve is $y = e^{x-1}$, which matches Option 4.

**Why Other Options Are Incorrect:**

- **Option 1: $x^2 + y^2 = 1$**
  - This represents a circle, and the slope of the tangent is not proportional to the ordinate. The normal at $P(1,1)$ would not have a slope of $-1$.

- **Option 2: $y^2 = x$**
  - This is a parabola, and the slope of the tangent is not proportional to the ordinate. The normal at $P(1,1)$ would not match the given equation.

- **Option 3: $(y-1)^2 = (x-1)$**
  - This is another form of a parabola, and similarly, the slope of the tangent is not proportional to the ordinate. The normal at $P(1,1)$ would not match the given equation.

Thus, Option 4 is the only equation that satisfies all the given conditions, making it the correct answer.