Skip to content
Question 88 of 96

Q.(a) Find the area of the region bounded by the ellipse x2a2+y2b2=1\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 OR

(b) Find the vertex, focus and equation of the directrix of the parabola y2−4y−8x+12=0y^2-4y-8x+12=0.
Puducherry TnboardTamil Nadu HSC (DGE) Board 2024Subjective· 5mImportance★★★★★
92% · 88/96 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

(a) Integrates the upper half of the ellipse over one quadrant and multiplies by symmetry to get the classic πab\pi ab area formula; (b) completes the square to bring the parabola to standard form and reads off its vertex, focus, and directrix. Both alternatives answered below.

(a) Area of the ellipse x2a2+y2b2=1\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1

1. First-quadrant area. Solving for y≥0y\ge0: y=b1−x2a2y=b\sqrt{1-\dfrac{x^2}{a^2}}. The area under this curve from x=0x=0 to x=ax=a is

∫0ab1−x2a2 dx=ba∫0aa2−x2 dx\displaystyle\int_0^a b\sqrt{1-\dfrac{x^2}{a^2}}\,dx=\dfrac ba\int_0^a\sqrt{a^2-x^2}\,dx

2. Standard integral. ∫0aa2−x2 dx=[x2a2−x2+a22sin⁡−1xa]0a=0+a22⋅π2=πa24\displaystyle\int_0^a\sqrt{a^2-x^2}\,dx=\left[\dfrac x2\sqrt{a^2-x^2}+\dfrac{a^2}2\sin^{-1}\dfrac xa\right]_0^a=0+\dfrac{a^2}2\cdot\dfrac\pi2=\dfrac{\pi a^2}4.

3. First-quadrant area =ba⋅πa24=πab4=\dfrac ba\cdot\dfrac{\pi a^2}4=\dfrac{\pi ab}4.

4. Full ellipse. By symmetry across both axes, total area =4×πab4=πab=4\times\dfrac{\pi ab}4=\pi ab.

(b) Parabola y2−4y−8x+12=0y^2-4y-8x+12=0

1. Complete the square in yy. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.