The curve `y=ax^2+bx +c` passes through the point `(1,2)` and its tangent at origin is the line `y=x`. The area bounded by the curve, the ordinate of the curve at minima and the tangent line is

Options

A. `1/24`
B. `1/12`
C. `1/8`
D. `1/6`

Answer

To solve the problem step by step, we will follow the reasoning provided in the video transcript while ensuring clarity in each step.

### Step 1: Determine the coefficients of the quadratic equation
The curve is given by the equation:
\[ y = ax^2 + bx + c \]

We know that the curve passes through the point (1, 2). Thus, substituting \( x = 1 \) and \( y = 2 \) into the equation gives:
\[ 2 = a(1)^2 + b(1) + c \]
\[ 2 = a + b + c \quad \text{(1)} \]

Additionally, since the tangent at the origin (0,0) is the line \( y = x \), the slope of the tangent at the origin is 1. The derivative of the curve at \( x = 0 \) is given by:
\[ \frac{dy}{dx} = 2ax + b \]
At \( x = 0 \):
\[ \frac{dy}{dx} = b = 1 \quad \text{(2)} \]

Since the curve passes through the origin, substituting \( x = 0 \) and \( y = 0 \) gives:
\[ 0 = a(0)^2 + b(0) + c \]
This implies:
\[ c = 0 \quad \text{(3)} \]

### Step 2: Substitute known values into the equations
From equations (2) and (3), we know:
\[ b = 1 \]
\[ c = 0 \]

Substituting these values into equation (1):
\[ 2 = a + 1 + 0 \]
Thus:
\[ a = 1 \]

### Step 3: Write the final equation of the curve
Now we have determined:
\[ a = 1, \quad b = 1, \quad c = 0 \]
So the equation of the curve becomes:
\[ y = x^2 + x \]

### Step 4: Find the vertex (minima) of the curve
The x-coordinate of the vertex of a parabola given by \( y = ax^2 + bx + c \) is found using:
\[ x = -\frac{b}{2a} \]
Substituting \( a = 1 \) and \( b = 1 \):
\[ x = -\frac{1}{2(1)} = -\frac{1}{2} \]

To find the y-coordinate at this point, substitute \( x = -\frac{1}{2} \) into the equation:
\[ y = \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right) \]
\[ y = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4} \]
Thus, the vertex (minima) is at:
\[ \left(-\frac{1}{2}, -\frac{1}{4}\right) \]

### Step 5: Set up the area calculation
We need to find the area bounded by the curve \( y = x^2 + x \), the tangent line \( y = x \), and the ordinate at the minima. The area can be calculated as the area between the curve and the line from \( x = -1 \) to \( x = 0 \).

### Step 6: Calculate the area between the curve and the line
The area \( A \) can be calculated using the integral:
\[ A = \int_{-1}^{0} \left( (x) - (x^2 + x) \right) dx \]
\[ = \int_{-1}^{0} (-x^2) dx \]
\[ = -\left[ \frac{x^3}{3} \right]_{-1}^{0} \]
\[ = -\left( 0 - \left(-\frac{1}{3}\right) \right) = \frac{1}{3} \]

### Step 7: Final area calculation
Now, we need to find the area of the triangle formed by the points (0,0), (-1, -1), and (-1, -1/4):
The base of the triangle is 1 (from x = -1 to x = 0) and the height is \( \frac{1}{4} \):
\[ \text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times \frac{1}{4} = \frac{1}{8} \]

### Step 8: Total area bounded
The total area bounded by the curve, the tangent line, and the ordinate at the minima is:
\[ A_{total} = A_{curve} - A_{triangle} \]
\[ A_{total} = \frac{1}{3} - \frac{1}{8} \]

Finding a common denominator (24):
\[ = \frac{8}{24} - \frac{3}{24} = \frac{5}{24} \]

### Final Answer
The area bounded by the curve, the ordinate of the curve at minima, and the tangent line is:
\[ \frac{5}{24} \text{ square units} \]

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