The origin of the coordinate plane is taken at the vertex of the parabolic reflector in such a way that the axis of the reflector is along the positive x-axis.
This can be diagrammatically represented as

The equation of the parabola is of the form y2 = 4ax (as it is opening to the right). Since the parabola passes through point A (5, 10),
\(10^2 = 4a(5)\)
\(⇒ 100 = 20a\)
\(⇒ a = \frac{100}{20} = 5\)
Therefore, the focus of the parabola is (a, 0) = (5, 0), which is the mid-point of the diameter.
Hence, the focus of the reflector is at the mid-point of the diameter.
Let \( y^2 = 12x \) be the parabola and \( S \) its focus. Let \( PQ \) be a focal chord of the parabola such that \( (SP)(SQ) = \frac{147}{4} \). Let \( C \) be the circle described by taking \( PQ \) as a diameter. If the equation of the circle \( C \) is: \[ 64x^2 + 64y^2 - \alpha x - 64\sqrt{3}y = \beta, \] then \( \beta - \alpha \) is equal to:
If \( x^2 = -16y \) is an equation of a parabola, then:
(A) Directrix is \( y = 4 \)
(B) Directrix is \( x = 4 \)
(C) Co-ordinates of focus are \( (0, -4) \)
(D) Co-ordinates of focus are \( (-4, 0) \)
(E) Length of latus rectum is 16
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:
Find the mean deviation about the mean for the data 38, 70, 48, 40, 42, 55, 63, 46, 54, 44.
Parabola is defined as the locus of points equidistant from a fixed point (called focus) and a fixed-line (called directrix).

=> MP2 = PS2
=> MP2 = PS2
So, (b + y)2 = (y - b)2 + x2