BITSAT 2011
- Exam
- BITSAT
- Year
- 2011
- Subject
- Mathematics
- Chapter
- Ellipse
BITSAT 2011 · Mathematics · Ellipse
The equation of the ellipse with focus at (( ± 5,0)) and...
The equation of the ellipse with focus at (( \pm 5,0)) and (x=\frac{36}{5}) as one directrix is
Options
- A(\frac{x^2}{36}+\frac{y^2}{25}=1)
- B(\frac{x^2}{36}+\frac{y^2}{11}=1)
- C(\frac{x^2}{25}+\frac{y^2}{11}=1)
- DNone of these
Correct answer
B. (\frac{x^2}{36}+\frac{y^2}{11}=1)
Step-by-step solution
We have ae [Since focus is (( \pm \mathrm{ae}, 0))]
and (\frac{\mathrm{a}}{\mathrm{e}}=\frac{36}{5} \quad\left[\right.) since directrix is (\left.\mathrm{x}= \pm \frac{\mathrm{a}}{\mathrm{e}}\right])
On solving we get (\mathrm{a}=6)
(\text{ and } \mathrm{e}=\frac{5}{6} \Rightarrow \mathrm{b}^2=\mathrm{a}^2\left(1-\mathrm{e}^2\right)=36\left(1-\frac{25}{36}\right)=11)
Thus, the required equation of the ellipse is
(\frac{x^2}{36}+\frac{y^2}{11}=1)