JEE Main201815 Apr 2018Evening ShiftMathematicsHyperbolaActual
A normal to the hyperbola, 4 x^2-9 y^2=36 meets the co-ordinate axes x and y at A and B , respectively. If the parallelogram O A B P(O being the origin) is formed, then the locus of P is
Options
- A4 x^2-9 y^2=121
- B4 x^2+9 y^2=121
- C9 x^2-4 y^2=169
- D9 x^2+4 y^2=169
Correct answer
C. 9 x^2-4 y^2=169
Step-by-step solution
Given, 4 x^2-9 y^2=36 After differentiating w.r.t. x , we get aligned & 4.2.x-9.2.y. d y d x =0 & Slope of tangent = d y d x = 4 x 9 y aligned So, slope of normal = -9 y 4 x Now, equation of normal at point (x₀, y₀ ) is given by y-y₀= -9 y₀ 4 x₀ (x-x₀ ) As normal intersects X axis at A , Then A ( 13 x₀ 9 , 0 ) and B (0, 13 y₀ 4 ) As O A B P is a parallelogram midpoint of O B (0, 13 y₀ 8 ) Midpoint of A P So, P(x, y) ( -13 x₀ 9 , 13 y₀ 4 ) (x₀, y₀ ) lies on hyperbola, therefore 4 (x₀ )^2-9 (y₀ )^2=36 From equation (