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Ellipse question

2008 · Shift 0 · Q45
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Ellipse question

2008 · Shift 0 · Q45

JEE MainMathematicsEllipseMCQ+4 / −1
A focus of an ellipse is at the origin. The directrix is the line x=4x=4x=4 and the eccentricity is 12{{1 \over 2}}21​. Then the length of the semi-major axis is :
  1. A
    83{{8 \over 3}}38​
  2. B
    23{{2 \over 3}}32​
  3. C
    43{{4 \over 3}}34​
  4. D
    53{{5 \over 3}}35​
View written solutionFree

Correct answer: A

  1. Use the focus-directrix definition of a conic

For any point P(x,y)P(x,y)P(x,y) on the ellipse,

distance from P to focusdistance from P to directrix=e=12.\frac{\text{distance from }P\text{ to focus}}{\text{distance from }P\text{ to directrix}} = e = \frac12.distance from P to directrixdistance from P to focus​=e=21​.

Given:

  • Focus at (0,0)(0,0)(0,0)
  • Directrix: x=4x=4x=4
  • Eccentricity: e=12e=\frac12e=21​

Since e<1e<1e<1, the conic is an ellipse.

  1. Find the vertex on the axis

The focus and directrix are both aligned along the xxx-axis, so the major axis lies along the xxx-axis.

Let the nearer vertex be at (x,0)(x,0)(x,0), lying between the focus and the directrix. Then

  • Distance to focus =∣x∣=|x|=∣x∣
  • Distance to directrix =∣4−x∣=4−x=|4-x|=4-x=∣4−x∣=4−x (since this point lies left of x=4x=4x=4)

Using the definition,

x4−x=12.\frac{x}{4-x}=\frac12.4−xx​=21​.

So,

2x=4−x2x=4-x2x=4−x 3x=43x=43x=4 x=43.x=\frac43.x=34​.

Thus one vertex is at (43,0)\left(\frac43,0\right)(34​,0).

  1. Relate this to the standard ellipse parameters

For an ellipse centered at (h,0)(h,0)(h,0) with major axis along xxx-axis:

  • focus positions are h±ch\pm ch±c
  • vertices are h±ah\pm ah±a
  • eccentricity is
e=ca.e=\frac{c}{a}.e=ac​.

Here the focus is at 000, and the right directrix is x=4x=4x=4. For such an ellipse,

directrix:x=h+ae.\text{directrix} : x=h+\frac{a}{e}.directrix:x=h+ea​.

Also,

0=h+c=h+aequad⇒h=−ae.0=h+c=h+ae quad\Rightarrow\quad h=-ae.0=h+c=h+aequad⇒h=−ae.

Hence,

4=h+ae=−ae+ae.4=h+\frac{a}{e}=-ae+\frac{a}{e}.4=h+ea​=−ae+ea​.

Substitute e=12e=\frac12e=21​:

4=−a(12)+a(2)=3a2.4=-a\left(\frac12\right)+a\left(2\right)=\frac{3a}{2}.4=−a(21​)+a(2)=23a​.

Therefore,

a=83.a=\frac{8}{3}.a=38​.
  1. Answer

The length of the semi-major axis is

83.\boxed{\frac{8}{3}}.38​​.

So the correct option is A.

  1. Comparison with stored answer

Stored correct answer: A

Our derived answer is also A, so they agree.

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