Let the focal chord PQ of the parabola $ y^2 = 4x $ make an angle of $ 60^\circ $ with the positive x-axis, where P lies in the first quadrant. If the circle, whose one diameter is PS, $ S $ being the focus of the parabola, touches the y-axis at the point $ (0, \alpha) $, then $ 5\alpha^2 $ is equal to:
Given the parabola \( y^2 = 4x \) and the geometry of the focal chord: \[ P( \sqrt{3}, 2\sqrt{3} ), S(1,0) \] The equation of the circle: \[ (x - 1)^2 + (y - 0)^2 = \left( \frac{\sqrt{3}}{2} \right)^2 \] The point where the circle touches the y-axis is found to be \( (0, \alpha) \), and substituting this into the equation gives \( 5\alpha^2 = 15 \).
Two parabolas have the same focus $(4, 3)$ and their directrices are the $x$-axis and the $y$-axis, respectively. If these parabolas intersect at the points $A$ and $B$, then $(AB)^2$ is equal to:
The term independent of $ x $ in the expansion of $$ \left( \frac{x + 1}{x^{3/2} + 1 - \sqrt{x}} \cdot \frac{x + 1}{x - \sqrt{x}} \right)^{10} $$ for $ x>1 $ is: