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Ellipse question

2019 · 12 Apr · Shift 2 · Q37
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Ellipse question

2019 · 12 Apr · Shift 2 · Q37

JEE MainMathematicsEllipseMCQ+4 / −1
An ellipse, with foci at (0, 2) and (0, –2) and minor axis of length 4, passes through which of the following points?
  1. A
    (2,2)\left( {2,\sqrt 2 } \right)(2,2​)
  2. B
    (2,22)\left( {2,2\sqrt 2 } \right)(2,22​)
  3. C
    (2,2)\left( {\sqrt 2 ,2} \right)(2​,2)
  4. D
    (1,22)\left( {1,2\sqrt 2 } \right)(1,22​)
View written solutionFree

Correct answer: C

  1. Identify the ellipse parameters

The foci are at (0,2)(0,2)(0,2) and (0,−2)(0,-2)(0,−2), so the center is at (0,0)(0,0)(0,0) and the major axis is along the yyy-axis.

Hence the standard form is

x2b2+y2a2=1,\frac{x^2}{b^2}+\frac{y^2}{a^2}=1,b2x2​+a2y2​=1,

with c=2.c=2.c=2.

Also, the minor axis length is 444, so 2b=4  ⟹  b=2  ⟹  b2=4.2b=4 \implies b=2 \implies b^2=4.2b=4⟹b=2⟹b2=4.

For an ellipse, a2=b2+c2=4+4=8.a^2=b^2+c^2=4+4=8.a2=b2+c2=4+4=8.

So the equation of the ellipse is

x24+y28=1.\frac{x^2}{4}+\frac{y^2}{8}=1.4x2​+8y2​=1.
  1. Check each option

We substitute each point into

x24+y28=1.\frac{x^2}{4}+\frac{y^2}{8}=1.4x2​+8y2​=1.

Option A: (2,2)\left(2,\sqrt2\right)(2,2​)

224+(2)28=44+28=1+14=54≠1.\frac{2^2}{4}+\frac{(\sqrt2)^2}{8}=\frac{4}{4}+\frac{2}{8}=1+\frac14=\frac54 \ne 1.422​+8(2​)2​=44​+82​=1+41​=45​=1.

So, A does not lie on the ellipse.

Option B: (2,22)\left(2,2\sqrt2\right)(2,22​)

224+(22)28=44+88=1+1=2≠1.\frac{2^2}{4}+\frac{(2\sqrt2)^2}{8}=\frac{4}{4}+\frac{8}{8}=1+1=2 \ne 1.422​+8(22​)2​=44​+88​=1+1=2=1.

So, B does not lie on the ellipse.

Option C: (2,2)\left(\sqrt2,2\right)(2​,2)

(2)24+228=24+48=12+12=1.\frac{(\sqrt2)^2}{4}+\frac{2^2}{8}=\frac{2}{4}+\frac{4}{8}=\frac12+\frac12=1.4(2​)2​+822​=42​+84​=21​+21​=1.

So, C does lie on the ellipse.

Option D: (1,22)\left(1,2\sqrt2\right)(1,22​)

124+(22)28=14+88=14+1=54≠1.\frac{1^2}{4}+\frac{(2\sqrt2)^2}{8}=\frac14+\frac{8}{8}=\frac14+1=\frac54 \ne 1.412​+8(22​)2​=41​+88​=41​+1=45​=1.

So, D does not lie on the ellipse.

  1. Conclusion

The point that lies on the ellipse is

(2,2).\boxed{\left(\sqrt2,2\right)}.(2​,2)​.

So the correct option is C.

  1. Comparison with stored answer

Stored correct answer: C

This matches our derived answer.

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