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 \(=\) \(4a= 6.\)
\(f(x) = \begin{cases} x^2, & \quad 0≤x≤3\\ 3x, & \quad 3≤x≤10 \end{cases}\)
The relation g is defined by
\(g(x) = \begin{cases} x^2, & \quad 0≤x≤2\\ 3x, & \quad 2≤x≤10 \end{cases}\)
Show that f is a function and g is not a function.
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