If the focus of the parabola \[ (y - k)^2 = 4(x - h) \] always lies between the lines $x + y = 1$ and $x + y = 3$ then:

Question

An equilateral triangle OAB is inscribed in the parabola $y^2 = 4x$ with the vertex O at the vertex of the parabola. Then the minimum distance of the circle having AB as a diameter from the origin is

Answer 1

The standard form of the parabola is \[(y - k)^2 = 4(x - h)\]The focus is located at \[(h + 1, k)\]Given that the focus lies between the lines $x + y = 1$ and $x + y = 3$, substituting the focus coordinates into these inequalities yields:\[1<(h+1) + k<3\]This simplifies to:\[0<h + k<2\]Therefore, the range for $ h + k $ is:\[0<h + k<2\]

If the focus of the parabola \[ (y - k)^2 = 4(x - h) \] always lies between the lines \(x + y = 1\) and \(x + y = 3\) then:

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The Correct Option is A

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