Question

If A is the area in the first quadrant enclosed by the curve \(C: 2x^2 – y + 1 = 0\), the tangent to C at the point (1, 3) and the line x + y = 1, then the value of 60A is ________

Answer 1

Correct Answer: 16

Shaded Area Calculation 

The equation of the curve is given as:

\( y = 2x^2 + 1 \)

The tangent at the point \( P(1, 3) \) is:

\( y = 4x - 1 \)

Step 1: Solve the Integral

We compute the integral of the curve from \( x = 0 \) to \( x = 1 \):

\[ \int_0^1 (2x^2 + 1) \, dx = \left[ \frac{2x^3}{3} + x \right]_0^1 \]

Evaluating the integral:

\[ = \left( \frac{2(1)^3}{3} + (1) \right) - \left( \frac{2(0)^3}{3} + 0 \right) = \frac{2}{3} + 1 = \frac{5}{3} \]

Step 2: Compute the Areas of the Triangles

1. Area of Triangle \( \triangle QOT \)

The area of \( \triangle QOT \) is:

\[ \text{Area} = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot 1 \cdot 1 = \frac{1}{2} \]

2. Area of Triangle \( \triangle PQR \)

The vertices of \( \triangle PQR \) are \( P(1,3), Q(1,0), R\left( \frac{1}{4}, 0 \right) \).

We compute the area using the formula for the area of a triangle with given vertices:

\[ \text{Area} = \frac{1}{2} \left| 1(0-0) + 1(0-3) + \frac{1}{4}(3-0) \right| = \frac{1}{2} \left| 0 - 3 + \frac{3}{4} \right| = \frac{1}{2} \cdot \frac{9}{4} = \frac{9}{8} \]

3. Area of Triangle \( \triangle QRS \)

The vertices of \( \triangle QRS \) are \( Q(1,0), R\left( \frac{1}{4}, 0 \right), S\left( \frac{2}{5}, \frac{13}{5} \right) \).

We compute the area similarly:

\[ \text{Area} = \frac{1}{2} \left| 1(0 - \frac{13}{5}) + \frac{1}{4}(\frac{13}{5} - 0) + \frac{2}{5}(0 - 0) \right| = \frac{1}{2} \left| -\frac{13}{5} + \frac{13}{20} \right| \]

Simplifying the expression:

\[ = \frac{1}{2} \cdot \frac{39}{40} = \frac{39}{40} \]

Step 3: Combine the Results

Now, we combine all the computed areas:

\[ A = \frac{5}{3} - \frac{1}{2} - \frac{9}{8} + \frac{9}{40} \]

To simplify, we find the common denominator (120):

\[ A = \frac{200}{120} - \frac{60}{120} - \frac{135}{120} + \frac{27}{120} = \frac{200 - 60 - 135 + 27}{120} = \frac{32}{120} = \frac{8}{30} = \frac{16}{60} \]

Final Answer

The shaded area is \( A = \frac{16}{60} \).