Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
AP EAMCET201824 Apr 2018Morning ShiftMathematicsHyperbolaActual

If the equation of one asymptote of the hyperbola 14 x^2+38 x y+20 y^2+x-7 y-91=0 is 7 x+5 y-3=0 , then the other asymptote is

Options

  1. A2 x-4 y+1=0
  2. B2 x+4 y+1=0
  3. C2 x-4 y-1=0
  4. D2 x+4 y-1=0

Correct answer

B. 2 x+4 y+1=0

Step-by-step solution

Given hyperbola, On factorising 14 x^2+38 x y+20 y^2 , we get =(7 x+5 y)(2 x+4 y) One of the asymptote is 7 x+5 y-3=0 Then, let other asymptote is 2 x+4 y+k=0 So, on combining On equating the coefficient of x from Eqs. (i) and (ii), we get 7 k-6=1 k=1 So, other asymptote is, 2 x+4 y+1=0 .

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