Find the coordinates of the focus, the axis of the parabola, the equation of directrix, and the length of the latus rectum for \(x^2 = 6y\)
The given equation is :\(x^2= 6y\)
Here, the coefficient of \(y\) is positive.
Hence, the parabola opens upwards.
On comparing this equation with \(x^2= 4ay\), we obtain
\(4a = 6\) \(x^2 \)\(\)
\(a = 6/4\)
\(= 3/2\)
∴Coordinates of the focus = \((0, a)\) \(=(0, 3/2) \)
Since the given equation involves \( x^2 \), the axis of the parabola is the y-axis.
Equation of directrix, \(y =-a\) i.e, \(y = -3/2\)
Length of the latus rectum \(=\) \(4a= 6.\)
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