PARABOLA(BASIC: BOARD)

The focal length is found by equating the general expression for y

$y=x^2/(4p)$

and our particular example:

$y=x^2/2$

So we have:

$x^2/(4p)=x^2/2$

This gives $p = 0.5$.

So the focus will be at $(0, 0.5)$ and the directrix is the line $y = -0.5$.

Our curve is as follows:

In this case, we have the following graph:

After sketching, we can see that the equation required is in the following form, since we have a horizontal axis:

$y^2 = 4px$

Since $p = -2$ (from the question), we can directly write the equation of the parabola:

$y^2 = -8x$

The curve must have the following orientation, since we know it has horizontal axis and passes through $(2, -1)$:

So we need to use the general formula for a parabola with horizontal axis:

$y^2= 4px$

We need to find `p`. We know the curve goes through $(2, -1)$, so we substitute:

$(-1)^2= 4(p)(2) $

→ `1 = 8p`

→ $p = 1/8$.

So the required equation is $y^2=x/2$.

Ans: 3y-4=0 and x=0. Compare with $x^2=4ay$>

Solution: 

The given parabola is y\(^{2}\) = 12x.

Now, let (k, 2k) be the co-ordinates of the required point (k ≠ 0).

Since the point lies (k, 2k) on the parabola y\(^{2}\) = 12x,

Therefore, we get,

 (2k)\(^{2}\) = 12k

⇒ 4k\(^{2}\) = 12k     

⇒ k = 3 (Since, k ≠ 0, ).

Therefore, the co-ordinates of the required point are (3, 6).