JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Let \(f(x)=x^2+9, g(x)=\frac{x}{x-9}\) and \(\mathrm{a}=\mathrm{fog}(10), \mathrm{b}=\operatorname{gof}(3)\). If \(\mathrm{e}\) and \(1\) denote the eccentricity and the length of the latus rectum of the ellipse \(\frac{x^2}{a}+\frac{y^2}{b}=1\), then \(8 e^2+1^2\) is equal to.
- A \(16\)
- B \(8\)
- C \(6\)
- D \(12\)
Answer & Solution
Correct Answer
(B) \(8\)
Step-by-step Solution
Detailed explanation
\( f(x)=x^2+9 \quad g(x)=\frac{x}{x-9} \) \( a=f(g(10))=f\left(\frac{10}{10-9}\right) \) \( =f(10)=109 \) \( b=g(f(3))=g(9+9) \) \( =g(18)=\frac{18}{9}=2 \) \( E: \frac{x^2}{109}+\frac{y^2}{2}=1 \) \( e^2=1-\frac{2}{109}=\frac{107}{109} \)…
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