Area Under the Curve
Area Enclosed by Parabola, Tangent and Line
nta_pyq_2023_apr
Grade 12
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................
Step-by-Step Solution
Key Concept: The parabola is $y=2x^2+1$. Tangent at $(1,3)$: $y=4x-1$. Find intersections of tangent with parabola and with $x+y=1$ to form the enclosed triangular-like region.
Tangent $y=4x-1$ meets $x+y=1$ at $(\frac{1}{2},\frac{1}{2})$ and $y=2x^2+1$ at $(1,3)$ and... Integrating the enclosed area gives $A=\frac{4}{15}$. $60A=16$.
Correct Answer: 16