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JEE Main202330 Jan 2023Evening ShiftMathematicsParabolaActual

The parabolas : a x 2 + 2 b x + c y = 0 and d 2 + 2 e x + f y = 0 intersect on the line y = 1 . If a , b , c , d , e , f are positive real numbers and a , b , c are in G . P . , then

Options

  1. Ad , e , f are in A.P.
  2. Bd a , e b , f c are in G.P.
  3. Cd a , e b , f c are in A.P.
  4. Dd , e , f are in G.P.

Correct answer

C. d a , e b , f c are in A.P.

Step-by-step solution

Given, a x 2 + 2 b x + c y = 0 and d 2 + 2 e x + f y = 0 intersect on the line y = 1 , And a , b , c are in G.P. So, b 2 = a c Now putting the value of b in a x 2 + 2 b x + c y = 0 and taking y = 1 we get, a x 2 + 2 b x + c = 0 ⇒ ax 2 + 2 ac x + c = 0   ∵ b 2 = ac ⇒ ( x a + c ) 2 = 0 ⇒ x = - c a and x 2 = c a   . . . . . . . 1 Now, putting the value of x   &   x 2 in d x 2 + 2 e x + f y = 0 and taking y = 1 we get, ⇒ d c a + 2 e - c a + f = 0 ⇒ d c a + f =

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