The tangent to the parabola y² = 4x at the point where it intersects the circle x² + y² = 5 in the first quadrant, passes through the point :

The shortest distance between the line y = x and the curve y² = x – 2 is :

The equation of a tangent to the parabola, x² = 8y, which makes an angle $$\theta $$ with the positive directions of x-axis, is :

Let P(4, –4) and Q(9, 6) be two points on the parabola, y² = 4x and let x be any point on the arc POQ of this parabola, where O is the vertex of this parabola, such that the area of $$\Delta $$PXQ is maximum. Then this maximum area (in sq. units) is :