AP EAMCET201920 Apr 2019Morning ShiftMathematicsParabolaActual
The locus of the points of intersection of perpendicular normals to the parabola (y^2=4 a x ) is
Options
- A(y^2-2 a x+a^2=0 )
- B(y^2+a x+2 a^2=0 )
- C(y^2-a x+2 a^2=0 )
- D(y^2-a x+3 a^2=0 )
Correct answer
D. (y^2-a x+3 a^2=0 )
Step-by-step solution
Let the equation of normal having slope ' (m ) ' to the parabola (y^2=4 a x ) is (y=m x-2 a m-a m^3 ) and passes through ((h, k) ), then (k=m h-2 a m-a m^3 ) is cubic equation in ' (m ) '. Let having roots (m₁, m₂ ) and (m₃ ). so, (m₁ m₂ m₃=- k a ) and (m₁ m₂=-1 ) due to perpendicular normals. ( aligned & so m₃= k a & k= k a h-2 a k a -a ( k a )^3 & I = h a -2- k^2 a^2 & aligned ) ( k^2 a^2 = h a -3 k^2=a(h-3 a) ) On taking locus ((h, k) ), we get (y^2-a x+3 a^2=0 ) Hence, option (4) is correct.