A curve which passes through (1,2) and whose sub-normal at every point is 2, is

To find the curve that passes through the point (1,2) with the property that the sub-normal at every point is 2, we can follow these steps:

1. ### Break Down the Problem

• The sub-normal (the length of the projection of the tangent onto the x-axis) is given by the formula: $\text{sub-normal} = \frac{y}{y'}$
• We need to set the equation for the sub-normal equal to 2, leading to: $\frac{y}{y'} = 2 \implies y = 2y'$

2. ### Relevant Concepts

• Recall that $y' = \frac{dy}{dx}$.
• From the equation $y = 2y'$, we can express $y'$ as: $y' = \frac{y}{2}$

3. ### Analysis and Detail

• We substitute $y' = \frac{dy}{dx}$: $\frac{dy}{dx} = \frac{y}{2}$
• This is a first-order differential equation and can be solved using separation of variables: $\frac{dy}{y} = \frac{1}{2}dx$
• Integrate both sides: $\int \frac{dy}{y} = \int \frac{1}{2}dx$ $\ln |y| = \frac{1}{2}x + C$
• Exponentiating both sides gives: $y = e^{\frac{1}{2}x + C} = e^{\frac{1}{2}x} \cdot e^C$
• If we let $e^C = k$, where $k$ is a constant, we have: $y = k e^{\frac{1}{2}x}$

4. ### Verify and Summarize

• The curve passes through the point (1,2). By substituting $x = 1$ and $y = 2$ into the equation: $2 = k e^{\frac{1}{2} \cdot 1} \implies 2 = k e^{\frac{1}{2}} \implies k = \frac{2}{e^{\frac{1}{2}}} = 2e^{-\frac{1}{2}}$
• Therefore, the equation becomes: $y = 2e^{-\frac{1}{2}} e^{\frac{1}{2}x} = 2e^{\frac{1}{2}(x-1)}$

Final Answer

The equation of the curve is: $y = 2e^{\frac{1}{2}(x-1)}$