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BITSAT 2015
Exam
BITSAT
Year
2015
Subject
Mathematics
Chapter
Ellipse

BITSAT 2015 · Mathematics · Ellipse

The eccentricity of an ellipse, with its centre at the origin, is 1...

The eccentricity of an ellipse, with its centre at the origin, is 1/21 / 2. If one of the directrices is x=4\mathrm{x}=4 then the equation of the ellipse is:

Options

  1. A4x2+3y2=14 x^{2}+3 y^{2}=1
  2. B3x2+4y2=123 x^{2}+4 y^{2}=12
  3. C4x2+3y2=124 x^{2}+3 y^{2}=12
  4. D3x2+4y2=13 x^{2}+4 y^{2}=1

Correct answer

B. 3x2+4y2=123 x^{2}+4 y^{2}=12

Step-by-step solution

e=12.\mathrm{e}=\frac{1}{2}. Directrix ,x=ae=4∴a=4×12=2∴b=21−14=3, \mathrm{x}=\frac{\mathrm{a}}{\mathrm{e}}=4 \therefore \mathrm{a}=4 \times \frac{1}{2}=2 \quad \therefore \mathrm{b}=2 \sqrt{1-\frac{1}{4}}=\sqrt{3} Equation of ellispe is x24+y23=1⇒3x2+4y2=12\frac{x^{2}}{4}+\frac{y^{2}}{3}=1 \Rightarrow 3 x^{2}+4 y^{2}=12
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