Find the coordinates of the focus, the axis of the parabola, the equation of directrix, and the length of the latus rectum for \(x^2 = - 16y\)
The given equation is y^2= 12x.
Here, the coefficient of x is positive.
Hence, the parabola opens towards the right.
On comparing this equation with \( y^2 = 4ax,\)
we obtain
\(4a= 12\)
\(⇒ a = 3\)
∴Coordinates of the focus=\(\)\( (a, 0) = (3, 0)\)
Since the given equation involves \(y^2\), the axis of the parabola is the x-axis.
Equation of direcctrix, \(x= -a \)
i.e., \(x = - 3 \)
\( x+ 3 = 0 \)
Length of the latus rectum \(= 4a = 12\)
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
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:
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