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Hyperbola question

2016 · 10 Apr · Shift 1 · Q34
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  5. /2016 · 10 Apr · Shift 1 · Q34

Hyperbola question

2016 · 10 Apr · Shift 1 · Q34

JEE MainMathematicsHyperbolaMCQ+4 / −1
A hyperbola whose transverse axis is along the major axis of the conic, x23+y24=4{{{x^2}} \over 3} + {{{y^2}} \over 4} = 43x2​+4y2​=4 and has vertices at the foci of this conic. If the eccentricity of the hyperbola is 32,{3 \over 2},23​, then which of the following points does NOT lie on it?
  1. A
    (0, 2)
  2. B
    (5,22)\left( {\sqrt 5 ,2\sqrt 2 } \right)(5​,22​)
  3. C
    (10,23)\left( {\sqrt {10} ,2\sqrt 3 } \right)(10​,23​)
  4. D
    (5,23)\left( {5,2\sqrt 3 } \right)(5,23​)
View written solutionFree

Correct answer: D

  1. Identify the given conic

The conic is

x23+y24=4.\frac{x^2}{3}+\frac{y^2}{4}=4.3x2​+4y2​=4.

Divide by 444:

x212+y216=1.\frac{x^2}{12}+\frac{y^2}{16}=1.12x2​+16y2​=1.

This is an ellipse with center at (0,0)(0,0)(0,0).

Since 16>1216>1216>12, its major axis is along the yyy-axis. So,

a2=16,b2=12.a^2=16,\quad b^2=12.a2=16,b2=12.

Hence the foci are

(0,±c),c2=a2−b2=16−12=4⇒c=2.(0,\pm c),\quad c^2=a^2-b^2=16-12=4 \Rightarrow c=2.(0,±c),c2=a2−b2=16−12=4⇒c=2.

Thus the foci of the ellipse are

(0,±2).(0,\pm 2).(0,±2).
  1. Form the required hyperbola

The hyperbola has its transverse axis along the major axis of the ellipse, so its transverse axis is also along the yyy-axis.

Also, its vertices are at the foci of the ellipse, i.e. at

(0,±2).(0,\pm 2).(0,±2).

Therefore for the hyperbola,

a=2⇒a2=4.a=2 \Rightarrow a^2=4.a=2⇒a2=4.

Since its transverse axis is along the yyy-axis, its equation is of the form

y2a2−x2b2=1.\frac{y^2}{a^2}-\frac{x^2}{b^2}=1.a2y2​−b2x2​=1.

So,

y24−x2b2=1.\frac{y^2}{4}-\frac{x^2}{b^2}=1.4y2​−b2x2​=1.
  1. Use the eccentricity

For a hyperbola,

e=ca,c2=a2+b2.e=\frac{c}{a}, \quad c^2=a^2+b^2.e=ac​,c2=a2+b2.

Given

e=32,a=2.e=\frac{3}{2}, \quad a=2.e=23​,a=2.

So,

c=ea=32⋅2=3.c=e a=\frac{3}{2}\cdot 2=3.c=ea=23​⋅2=3.

Hence,

c2=9=a2+b2=4+b2.c^2=9=a^2+b^2=4+b^2.c2=9=a2+b2=4+b2.

Thus,

b2=5.b^2=5.b2=5.

Therefore the hyperbola is

y24−x25=1.\frac{y^2}{4}-\frac{x^2}{5}=1.4y2​−5x2​=1.
  1. Check each option

We test whether each point satisfies

y24−x25=1.\frac{y^2}{4}-\frac{x^2}{5}=1.4y2​−5x2​=1.

Option A: (0,2)(0,2)(0,2)

224−025=44=1.\frac{2^2}{4}-\frac{0^2}{5}=\frac{4}{4}=1.422​−502​=44​=1.

So A lies on the hyperbola.

Option B: (5,22)\left(\sqrt5,2\sqrt2\right)(5​,22​)

(22)24−(5)25=84−55=2−1=1.\frac{(2\sqrt2)^2}{4}-\frac{(\sqrt5)^2}{5} =\frac{8}{4}-\frac{5}{5} =2-1=1.4(22​)2​−5(5​)2​=48​−55​=2−1=1.

So B lies on the hyperbola.

Option C: (10,23)\left(\sqrt{10},2\sqrt3\right)(10​,23​)

(23)24−(10)25=124−105=3−2=1.\frac{(2\sqrt3)^2}{4}-\frac{(\sqrt{10})^2}{5} =\frac{12}{4}-\frac{10}{5} =3-2=1.4(23​)2​−5(10​)2​=412​−510​=3−2=1.

So C lies on the hyperbola.

Option D: (5,23)\left(5,2\sqrt3\right)(5,23​)

(23)24−525=124−255=3−5=−2≠1.\frac{(2\sqrt3)^2}{4}-\frac{5^2}{5} =\frac{12}{4}-\frac{25}{5} =3-5=-2 \ne 1.4(23​)2​−552​=412​−525​=3−5=−2=1.

So D does not lie on the hyperbola.

  1. Conclusion

The point that does not lie on the hyperbola is

D (5,23).\boxed{\text{D } (5,2\sqrt3)}.D (5,23​)​.

This matches the stored correct answer.

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