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Ellipse question

2019 · 10 Jan · Shift 2 · Q33
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Ellipse question

2019 · 10 Jan · Shift 2 · Q33

JEE MainMathematicsEllipseMCQ+4 / −1
Let S = {(x,y)∈R2:y21+r−x21−r};re±1.\left\{ {\left( {x,y} \right) \in {R^2}:{{{y^2}} \over {1 + r}} - {{{x^2}} \over {1 - r}}} \right\};r e \pm 1.{(x,y)∈R2:1+ry2​−1−rx2​};re±1. Then S represents :
  1. A
    an ellipse whose eccentricity is 1r+1,{1 \over {\sqrt {r + 1} }},r+1​1​, where r > 1
  2. B
    an ellipse whose eccentricity is 2r+1,{2 \over {\sqrt {r + 1} }},r+1​2​, where 0 < r < 1
  3. C
    an ellipse whose eccentricity is 2r−1,{2 \over {\sqrt {r - 1} }},r−1​2​, where 0 < r < 1
  4. D
    an ellipse whose eccentricity is 2r+1\sqrt {{2 \over {r + 1}}}r+12​​, where r > 1
View written solutionFree

Correct answer: NONE OF THE OPTIONS, CORRECT RESULT: $E=\SQRT{\FRAC{2R}{R+1}}$ FOR $0<R<1$

The given set is intended to be interpreted as the conic

S={(x,y)∈R2: y21+r+x21−r=1, r≠±1}S=\left\{(x,y)\in \mathbb R^2:\ \frac{y^2}{1+r}+\frac{x^2}{1-r}=1,\ r\neq \pm 1\right\}S={(x,y)∈R2: 1+ry2​+1−rx2​=1, r=±1}

(or equivalently with the denominators interchanged). Since the options ask for an ellipse and its eccentricity, we determine when the equation represents an ellipse and then compute eee.

1. Condition for ellipse

A standard ellipse has the form

x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1a2x2​+b2y2​=1

with both denominators positive.

Here the denominators are 1+r1+r1+r and 1−r1-r1−r.

  • 1+r>0⇒r>−11+r>0 \Rightarrow r>-11+r>0⇒r>−1
  • 1−r>0⇒r<11-r>0 \Rightarrow r<11−r>0⇒r<1

So for both to be positive, we need

−1<r<1.-1<r<1.−1<r<1.

Among the given options, this means the relevant range is

0<r<1.0<r<1.0<r<1.

Hence any option with r>1r>1r>1 cannot represent an ellipse.

So options A and D are ruled out immediately.

2. Identify major and minor axes

For 0<r<10<r<10<r<1,

1+r>1−r.1+r>1-r.1+r>1−r.

Thus the larger denominator is 1+r1+r1+r, so this is the square of the semi-major axis:

a2=1+r,a^2=1+r,a2=1+r,

and the smaller denominator is

b2=1−r.b^2=1-r.b2=1−r.

3. Compute eccentricity

For an ellipse,

e=1−b2a2.e=\sqrt{1-\frac{b^2}{a^2}}.e=1−a2b2​​.

Substitute a2=1+ra^2=1+ra2=1+r and b2=1−rb^2=1-rb2=1−r:

e=1−1−r1+r=(1+r)−(1−r)1+r=2r1+r.e=\sqrt{1-\frac{1-r}{1+r}} =\sqrt{\frac{(1+r)-(1-r)}{1+r}} =\sqrt{\frac{2r}{1+r}}.e=1−1+r1−r​​=1+r(1+r)−(1−r)​​=1+r2r​​.

This does not match any option exactly as written.

However, if the intended equation was the common textbook form

x21+r+y21−r=1,\frac{x^2}{1+r}+\frac{y^2}{1-r}=1,1+rx2​+1−ry2​=1,

then the same computation still gives

e=2r1+r,0<r<1.e=\sqrt{\frac{2r}{1+r}}, \qquad 0<r<1.e=1+r2r​​,0<r<1.

So the options seem to contain a typographical error.

4. Compare with options

  • A: requires r>1r>1r>1 for ellipse — impossible.
  • B: e=2r+1e=\dfrac{2}{\sqrt{r+1}}e=r+1​2​ for 0<r<10<r<10<r<1 — not equal to 2r1+r\sqrt{\dfrac{2r}{1+r}}1+r2r​​.
  • C: also incorrect form and range.
  • D: requires r>1r>1r>1 for ellipse — impossible.

Therefore, none of the options is correct for the ellipse represented by the given equation.

5. Conclusion

The correct eccentricity should be

e=2rr+1for 0<r<1.\boxed{e=\sqrt{\frac{2r}{r+1}} \quad \text{for } 0<r<1.}e=r+12r​​for 0<r<1.​

So the stored answer D is not correct.

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