The area (in sq. units) of the part of circle x²+y²=169$x^2+y^2=169$ which is below the line 5x-y=13$5x-y=13$ is (πα)/(2β)-(65)/(2)+(α)/(β) ⁻¹((12)/(13))$\frac{\pi\alpha}{2\beta}-\frac{65}{2}+\frac{\alpha}{\beta}\sin^{-1}(\frac{12}{13})$ where α,β$\alpha,\beta$ are coprime numbers. Then α+β$\alpha+\beta$ is equal to
Numerical Answer Type:
Enter a numerical valueAnswer: 171 to 171+4 marks
Solution & Explanation
Related Formula
Standard Integral: ∫ √(a²-y²) dy = (y)/(2)√(a²-y²) + (a²)/(2) ⁻¹((y)/(a)) + C$$\text{Standard Integral: } \int \sqrt{a^2-y^2} dy = \frac{y}{2}\sqrt{a^2-y^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{y}{a}\right) + C$$Area of right triangle = (1)/(2) × base × height$$\text{Area of right triangle } = \frac{1}{2} \times \text{base} \times \text{height}$$
Core Logic
First, find the points of intersection between the circle x²+y²=169$x^2+y^2=169$ and the line 5x-y=13 ⇒ y = 5x-13$5x-y=13 \Rightarrow y = 5x-13$.
Substitute y$y$ into the circle equation:
The solutions are x=0$x=0$ and x=5$x=5$.
When x=0, y=-13$x=0, y=-13$. Point is (0, -13)$(0, -13)$.
When x=5, y=12$x=5, y=12$. Point is (5, 12)$(5, 12)$.
The required area is bounded below the line x = (y+13)/(5)$x = \frac{y+13}{5}$ and above the right-hand boundary of the circle x = √(169-y²)$x = \sqrt{169-y^2}$ across the y-axis boundaries [-13, 12]$[-13, 12]$.
Area Under Curves
Step 1: Setup Area Integral
Integrate with respect to y$y$ (from left to right curves, bounded horizontally):
Split the integral into two parts:
Part A (Circle): ∫₋₁₃¹² √(169-y²) dy$\int_{-13}^{12} \sqrt{169-y^2} dy$
Part B (Line): ∫₋₁₃¹² (y+13)/(5) dy$\int_{-13}^{12} \frac{y+13}{5} dy$
Comparing this exactly with the given format (πα)/(2β) - (65)/(2) + (α)/(β) ⁻¹((12)/(13))$\frac{\pi\alpha}{2\beta} - \frac{65}{2} + \frac{\alpha}{\beta}\sin^{-1}(\frac{12}{13})$:
We see that (α)/(β) = (169)/(2)$\frac{\alpha}{\beta} = \frac{169}{2}$.
Since 169 and 2 are coprime, α = 169$\alpha = 169$ and β = 2$\beta = 2$.
When evaluating line integrals forming a triangle with horizontal bounds, bypass algebraic integration and visually calculate (1)/(2) · b · h$\frac{1}{2} \cdot b \cdot h$. Here, base=25 along y-axis, height=5 along x-axis, area = 125/2$125/2$. Instantly saves integration time.
Chapter Mix
Class 12 Mathematics: Application of Integrals
Class 11 Mathematics: Straight Lines
Q5jee_main_2024_31_jan_morningArea bounded by Parabolas and Inequalities
The area of the region (x,y): y² ≤ 4x, x < 4, (xy(x - 1)(x - 2))/((x - 3)(x - 4)) > 0, x ≠ 3$\left\{(x,y): y^2 \le 4x, x < 4, \frac{xy(x - 1)(x - 2)}{(x - 3)(x - 4)} > 0, x \neq 3\right\}$ is
A.(16)/(3)$$\frac{16}{3}$$
B.(64)/(3)$$\frac{64}{3}$$
C.(8)/(3)$$\frac{8}{3}$$
D.(32)/(3)$$\frac{32}{3}$$
Solution
Core Logic
Given y² ≤ 4x$y^2 \le 4x$ and x < 4$x < 4$.
Analyze the inequality (xy(x-1)(x-2))/((x-3)(x-4)) > 0$\frac{xy(x-1)(x-2)}{(x-3)(x-4)} > 0$ considering y > 0$y > 0$ and y < 0$y < 0$ separately.
Area bounded by Parabolas and Inequalities diagram for Q5 - JEE Main 2024 Morning
Step 1: Case I (y > 0)
If y > 0$y > 0$, the inequality reduces to (x(x-1)(x-2))/((x-3)(x-4)) > 0$\frac{x(x-1)(x-2)}{(x-3)(x-4)} > 0$.
Using wavy curve method and given x in (0, 4)$x \in (0, 4)$:
x in (0, 1) (2, 3)$x \in (0, 1) \cup (2, 3)$.
Step 2: Case II (y < 0)
If y < 0$y < 0$, the inequality reduces to (x(x-1)(x-2))/((x-3)(x-4)) < 0$\frac{x(x-1)(x-2)}{(x-3)(x-4)} < 0$.
Using wavy curve method and given x in (0, 4)$x \in (0, 4)$:
x in (1, 2) (3, 4)$x \in (1, 2) \cup (3, 4)$.
Step 3: Area Computation
Because the regions map perfectly without overlap in opposite quadrants relative to the x-axis, they form complete parabolic strips when combined:
Area = 2 ∫₀⁴ √(x) dx = 2 · (2)/(3)[x3/2]₀⁴ = (4)/(3) · 8 = (32)/(3)$= 2 \int_{0}^{4} \sqrt{x} dx = 2 \cdot \frac{2}{3}[x^{3/2}]_{0}^{4} = \frac{4}{3} \cdot 8 = \frac{32}{3}$.
Chapter Mix
Class 12 Maths: Area Under Curves
Class 11 Maths: Linear Inequalities
More Application of Integrals Questions — jee_main_2024_29_jan_morning
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
