Skip to content
Exercise 7.2 · Q31

Q.Find the

(i) lengths of the principal axes
(ii) co-ordinates of the foci
(iii) equations of directrics
(iv) length of the latus rectum
(v) distance between foci
(vi) distance between directrices of the ellipse: 2x2+6y2=62x^2 + 6y^2 = 6.
Maharashtra MsbshseTextbookSubjectiveImportance★★★★★est
21% · 31/151 Questions
✓ Free question

2x2+6y2=6⇒x23+y2=12x^2+6y^2=6 \Rightarrow \dfrac{x^2}{3}+y^2=1. So a2=3, b2=1⇒a=3, b=1a^2=3,\,b^2=1 \Rightarrow a=\sqrt3,\,b=1.

  1. Major axis =23=2\sqrt3; minor axis =2=2.
  2. e2=1−13=23⇒e=23=63e^2=1-\dfrac13=\dfrac23 \Rightarrow e=\sqrt{\dfrac23}=\dfrac{\sqrt6}{3}. Foci (±ae,0)=(±3⋅63,0)=(±2,0)(\pm ae,0)=\left(\pm\sqrt3\cdot\dfrac{\sqrt6}{3},0\right)=(\pm\sqrt2,0).
  3. Directrices: x=±ae=±322x=\pm\dfrac{a}{e}=\pm\dfrac{3\sqrt2}{2}.
  4. Latus rectum =2b2a=23=233=\dfrac{2b^2}{a}=\dfrac{2}{\sqrt3}=\dfrac{2\sqrt3}{3}.
  5. Distance between foci =2ae=22=2ae=2\sqrt2.
  6. Distance between directrices =2ae=32=\dfrac{2a}{e}=3\sqrt2.
    ✓Final answer

    (i) 23,22\sqrt3,2 (ii) (±2,0)(\pm\sqrt2,0) (iii) x=±322x=\pm\tfrac{3\sqrt2}{2} (iv) 233\tfrac{2\sqrt3}{3} (v) 222\sqrt2 (vi) 323\sqrt2.

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.