Question

Let minimum value of $f(x) = |x - a| + 2|x - 2|$ is $1 \ a > 2$. Consider an ellipse $\frac{x^2}{a^2} + \frac{y^2}{4} = 1$. Tangents are drawn from point P(1, 2) to the ellipse.

On the basis of above information answer the following:

If 'e' is eccentricity of given ellipse, then e can be -

• A. $\frac{\sqrt{3}}{2}$
• B. $\frac{\sqrt{5}}{3}$
• C. $\frac{1}{3}$
• D. None of these

Answer 1

Answer: (B)

$\frac{\sqrt{5}}{3}$

Answer 2

The minimum value of $f(x) = |x - a| + 2|x - 2|$ with $a > 2$ occurs at $x=2$. Minimum value $= f(2) = |2 - a| + 2|2 - 2| = |2 - a|$. Since $a > 2$, $|2 - a| = a - 2$. Given minimum value is 1, $a - 2 = 1 \implies a = 3$. The ellipse is $\frac{x^2}{a^2} + \frac{y^2}{4} = 1 \implies \frac{x^2}{3^2} + \frac{y^2}{4} = 1 \implies \frac{x^2}{9} + \frac{y^2}{4} = 1$. Semi-major axis $a' = 3$, semi-minor axis $b' = 2$. Eccentricity $e = \sqrt{1 - \frac{b'^2}{a'^2}} = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}$.