AP EAMCET · Maths · Ellipse
Let F and \(\mathrm{F}^1\) be the foci of the ellipse \(\frac{x^2}{4}+\frac{y^2}{b^2}=1(b \lt 2)\) and \(B\) is one end of the minor axis. If the area of the triangle \(\mathrm{FBF}^1\) is \(\sqrt{3}\) sq. units, then the eccentricity of the ellipse is
- A \(\frac{\sqrt{3}}{2}\) or \(\frac{1}{2}\)
- B \(\frac{1}{\sqrt{3}}\)
- C \(\frac{\sqrt{3}}{4}\) or \(\frac{1}{4}\)
- D \(\frac{3}{4}\) or \(\frac{1}{4}\)
Answer & Solution
Correct Answer
(A) \(\frac{\sqrt{3}}{2}\) or \(\frac{1}{2}\)
Step-by-step Solution
Detailed explanation
Given equation of ellipse is \(\frac{x^2}{4}+\frac{y^2}{b^2}=1\) and \((b \lt 2)\) so, \(F(c, 0), F^{\prime}(-c, 0)\) and \(B(0, b)\) Area of \(\triangle F B F^{\prime}=\frac{1}{2} \times 2 c \times b\) \(\sqrt{3}=b c \Rightarrow b^2 c^2=3 \Rightarrow c^2=\frac{3}{b^2}\) ....(i)…
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