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Ellipse: Minimum Value Take Ellipse Draw Tangents Point Ellipse

JEE Maths question with a full step-by-step solution.

Question

The minimum value of

f(x) = |x-a| + 2|x-2|

1

, and

a > 2

. Take the ellipse

\dfrac{x^{2}}{a^{2}} + \dfrac{y^{2}}{4} = 1

and draw tangents to it from the point

P(1,2)

. If this ellipse has eccentricity

e

, then a possible value of

e

\dfrac{\sqrt3}{2}

\dfrac{\sqrt5}{3}

correct

\dfrac13

DNone of these

Solution

Step 1:

a > 2

, so

f(x) = \begin{cases} a + 4 - 3x, & x < 2 \\ a - 4 + x, & 2 \le x \le a \\ 3x - a - 4, & x > a \end{cases}

The slope is

-3

x<2

and

+1

2 < x < a

, so

f

decreases up to

x=2

and increases after it. The minimum is at

x = 2

Step 2:

f(2) = |2-a| + 0 = |a-2|

Step 3:

|a-2| = 1 \ \Longrightarrow\ a = 3 \text{ or } a = 1

a > 2

, so

a = 3

. Step 4:

\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1

9 > 4

, so the semi-major axis is

3

along

x

and the semi-minor axis is

2

. Step 5:

e = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac59} = \frac{\sqrt5}{3}

Answer: (2).

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