A curve is such that the mid-point of the portion of the tangent intercepted between the point where the tangent is drawn and the point where the tangent meets the y-axis lies on the line `y=xdot` If the curve passes through `(1,0),` then the curve is (a) `( b ) (c)2y=( d ) x^(( e )2( f ))( g )-x (h)` (i) (b) `( j ) (k) y=( l ) x^(( m )2( n ))( o )-x (p)` (q) (c) `( d ) (e) y=x-( f ) x^(( g )2( h ))( i ) (j)` (k) (d) `( l ) (m) y=2(( n ) (o) x-( p ) x^(( q )2( r ))( s ) (t))( u )` (v)

Options

A. `2y=x^(2)-x`
B. `y=x^(2)-x`
C. `y=x-x^(2)`
D. `y=2(x-x^(2))`

Explanation

To solve the problem, let's follow the steps outlined in the video transcript and derive the equation of the curve step by step.

### Step 1: Identify the Points
We know that the tangent to the curve at a point \((x, y)\) meets the y-axis at a point \((0, y - x \frac{dy}{dx})\).

**Hint:** Remember that the y-intercept of a line can be found using the point-slope form of the line equation.

### Step 2: Midpoint Calculation
The midpoint of the segment intercepted by the tangent between the point where the tangent is drawn and where it meets the y-axis is given by:
\[ \left(\frac{x + 0}{2}, \frac{y + \left(y - x \frac{dy}{dx}\right)}{2}\right) = \left(\frac{x}{2}, \frac{2y - x \frac{dy}{dx}}{2}\right) \]

**Hint:** Use the midpoint formula \(\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\).

### Step 3: Condition Given in the Problem
According to the problem, this midpoint lies on the line \(y = x\). Therefore, we can set up the equation:
\[ \frac{2y - x \frac{dy}{dx}}{2} = \frac{x}{2} \]

**Hint:** Substitute the midpoint coordinates into the equation of the line \(y = x\).

### Step 4: Simplifying the Equation
Multiplying both sides by 2 to eliminate the fraction gives:
\[ 2y - x \frac{dy}{dx} = x \]
Rearranging this, we find:
\[ 2y - x = x \frac{dy}{dx} \]
Thus,
\[ \frac{dy}{dx} = \frac{2y - x}{x} \]

**Hint:** Isolate \(\frac{dy}{dx}\) to express it in terms of \(y\) and \(x\).

### Step 5: Substituting \(y\) in terms of \(x\)
Let \(v = \frac{y}{x}\), so \(y = vx\). Then, we can differentiate \(y\) with respect to \(x\):
\[ \frac{dy}{dx} = v + x \frac{dv}{dx} \]
Substituting this back into our equation gives:
\[ v + x \frac{dv}{dx} = \frac{2(vx) - x}{x} = 2v - 1 \]
This simplifies to:
\[ x \frac{dv}{dx} = 2v - 1 - v = v - 1 \]
Thus,
\[ \frac{dv}{dx} = \frac{v - 1}{x} \]

**Hint:** Use the method of separation of variables to solve this differential equation.

### Step 6: Solving the Differential Equation
Separating variables gives:
\[ \frac{dv}{v - 1} = \frac{dx}{x} \]
Integrating both sides:
\[ \int \frac{dv}{v - 1} = \int \frac{dx}{x} \]
This results in:
\[ \ln |v - 1| = \ln |x| + C \]
Exponentiating both sides yields:
\[ |v - 1| = k|x| \quad \text{(where \(k = e^C\))} \]
Thus,
\[ v - 1 = kx \quad \text{or} \quad v = kx + 1 \]

**Hint:** Remember to consider the constant of integration when integrating.

### Step 7: Replacing \(v\) Back
Substituting back \(v = \frac{y}{x}\):
\[ \frac{y}{x} = kx + 1 \implies y = kx^2 + x \]

### Step 8: Using the Initial Condition
Given that the curve passes through the point \((1, 0)\):
\[ 0 = k(1)^2 + 1 \implies k + 1 = 0 \implies k = -1 \]
Thus, the equation of the curve becomes:
\[ y = -x^2 + x \]
or rearranging gives:
\[ y = x - x^2 \]

**Final Answer:**
The curve is \(y = x - x^2\).