JEE Main 2019 — Hyperbola Question with Solution
From: JEE Main 2019 (Online) 11th January Morning Slot
Question
Equation of a common tangent to the parabola y2 = 4x and the hyperbola xy = 2 is :
Choose an option
Show full solutionCorrect option: D
Correct answer
Dx + 2y + 4 = 0
Step-by-step explanation
Let the equation of tangent to parabola
y2 = 4x be y = mx +
It is also a tangent to hyperbola xy = 2
x = 2
x2m + 2 = 0
D = 0 m =
So tangent is 2y + x + 4 = 0
y2 = 4x be y = mx +
It is also a tangent to hyperbola xy = 2
x = 2
x2m + 2 = 0
D = 0 m =
So tangent is 2y + x + 4 = 0
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