Question

A common tangent to the conic x^2=6y and 2x^2-4y^2=9 is

Answer 1

Answer:

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Answer 2

Step-by-step explanation: given equation x² = 6y  and 2x² -4y² = 9

4a = 6 ⇒ a = 3/2              and     x²/9/2  -  y²/9/4 = 1

equation of tangent in parametric form     x = y/t + at from here y = tx - 3/2 t²  -------------------- (1)

equation of tangent in hyperbola y = mx ±√a²m² -b²  from above a² = 9/2 and b² = 9/4      ⇒  y = mx±√9/2m² - 9/4 ------------------ (2) equation first and second represent common tangent , so compare these equation

m/t = -1/-1 = ±√9m²/2 - 9/4 compare these  m= t    and second

±√9m²/2 - 9/4 = -3/2 m²  squaring both side we get   m = ± 1 so  t = ± 1

put these values in equation  1 we get required common tangent

  y = x -3/2    and  y = -x -3/2