Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
AP EAMCET2015MathematicsHyperbola

The foci of the ellipse x^2 16 + y^2 b^2 =1 and the hyperbola x^2 144 - y^2 81 = 1 25 coincide. Then, the value of b^2 is

Options

  1. A5
  2. B7
  3. C9
  4. D1

Correct answer

B. 7

Step-by-step solution

Equation of ellipse, x^2 16 + y^2 b^2 =1 Eccentricity, e= 1- b^2 16 = 16-b^2 4 So, the focus will be ( 16-b^2 , 0 ) Also, the equation of hyperbola, x^2 144 - y^2 81 = 1 25 x^2 ( 144 25 ) - y^2 ( 81 25 ) =1 e= 1+ b^2 a^2 = 1+ 81 25 144 25 = 15 12 So, the focus will be ( 12 5 15 12 , 0 ) i.e., ( 3,0) On comparing the focus of ellipse with hyperbola, we get aligned & 16-b^2 =3 & 16-b^2=9 & b^2=7 aligned

Practice Hyperbola on Quantrex Academy →

More from Hyperbola

The difference between the distance of any point on the hyperbola from the two foci is 16 and the eccentricity is 2 . Then the equation of the hyperbola is 2026In the figure Statement-I: When > 0 , the section is hyperbola Statement-II: When > 90^ , the section is ellipse Which of the following is correct? 2026If the line y = 2x + is a tangent to the hyperbola 36x^2 - 25y^2 = 3600 , then = 2026If the line 4x + 3y = 7 touches the hyperbola x^2 - y^2 = 7 , then the sum of the co-ordinates of the point of contact is... 2026If the eccentricity of the hyperbola x^2 a^2 - y^2 b^2 =1 passing through the point (4,6) is 2, then the equation of the tangent to this hyperbola at (4,6) is 2025A hyperbola passes through the point P ( 2 , 3 ) and has foci at ( 2,0) . Then the point that lies on the tangent drawn to this hyperbola at P is 2025If is the angle subtended by a latus rectum at the centre of the hyperbola having eccentricity 2 7 - 3 , then = 2025The tangent drawn at an extremity (in the first quadrant) of latus rectum of the hyperbola x^2 4 - y^2 5 =1 meets the x -axis and y -axis at A and B respectively. If O is the origi 2025 Full Hyperbola list All AP EAMCET PYQs