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Hyperbola question

2019 · 11 Jan · Shift 2 · Q22
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  5. /2019 · 11 Jan · Shift 2 · Q22

Hyperbola question

2019 · 11 Jan · Shift 2 · Q22

JEE MainMathematicsHyperbolaMCQ+4 / −1
If a hyperbola has length of its conjugate axis equal to 5 and the distance between its foci is 13, then the eccentricity of the hyperbola is :
  1. A
    136{{13} \over 6}613​
  2. B
    2
  3. C
    1312{{13} \over 12}1213​
  4. D
    138{{13} \over 8}813​
View written solutionFree

Correct answer: C

  1. For a standard hyperbola, x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1a2x2​−b2y2​=1 the relevant quantities are:

    • Length of conjugate axis =2b= 2b=2b
    • Distance between foci =2c= 2c=2c
    • Relation: c2=a2+b2c^2=a^2+b^2c2=a2+b2
    • Eccentricity: e=cae=\frac{c}{a}e=ac​

  2. Given length of conjugate axis is 555: 2b=5  ⟹  b=522b=5 \implies b=\frac{5}{2}2b=5⟹b=25​

  3. Given distance between foci is 131313: 2c=13  ⟹  c=1322c=13 \implies c=\frac{13}{2}2c=13⟹c=213​

  4. Use the hyperbola relation: c2=a2+b2c^2=a^2+b^2c2=a2+b2 So, a2=c2−b2=(132)2−(52)2a^2=c^2-b^2=\left(\frac{13}{2}\right)^2-\left(\frac{5}{2}\right)^2a2=c2−b2=(213​)2−(25​)2 a2=1694−254=1444=36a^2=\frac{169}{4}-\frac{25}{4}=\frac{144}{4}=36a2=4169​−425​=4144​=36 a=6a=6a=6

  5. Now compute eccentricity: e=ca=13/26=1312e=\frac{c}{a}=\frac{13/2}{6}=\frac{13}{12}e=ac​=613/2​=1213​

  6. Therefore, the correct option is: 1312\boxed{\frac{13}{12}}1213​​ which is Option C.

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