Skip to main contentSkip to solution

The area of the smaller region bounded by ellipse x216+y29=1\dfrac{x^2}{16}+\dfrac{y^2}{9}=1 and the straight line 3x+4y=123x+4y=12 is

Solution

✅ Correct Option: 2

The line x4+y3=1\dfrac{x}{4}+\dfrac{y}{3} = 1 joins the ellipse's intercepts (4,0)(4,0) and (0,3)(0,3).

Smaller area == (quarter ellipse) −- (right triangle) =14π(4)(3)−12(4)(3)=3π−6= \dfrac{1}{4}\pi(4)(3) - \dfrac{1}{2}(4)(3) = 3\pi - 6 sq. units.

Keyboard Shortcuts

  • Left arrow: Previous question
  • Right arrow: Next question
  • S key: Jump to solution
  • Q key: Jump to question