Question

In case of parabola

y= x2-2x-3

Find:

a)Vertex:

b)Axis:

c)Focus:

d)Directrix:

e)Latus Rectum:

Answer 1

Values:

a)v(1,-4)

b)x=1

c)f(1,-3.75)

d)y=-4.25

e)latus rectum=0.5units

Explanation

$y = {x}^{2} - 2x - 3 \\ y = {(x)}^{2} + 2(x)( - 1) + {( - 1)}^{2} - 3 - {( - 1)}^{2} \\ y = {(x - 1)}^{2} - 4$

Now for the vertex, we know a parabola's lowest point will be at it's line of symmetry/axis.

$vertex \\ at \: x = 1 \\ y = {(1 - 1)}^{2} - 4 \\ y = - 4 \\ v(1, - 4)$

Now we already know it's axis will be at

$x = 1$

We can deduce that the above equation is a parabola with it's line of symmetry at x=1 and shifted down by 4 units. To find out the focus of a parabola, we compare it with the standard equation.

$4ay = {x}^{2} \\ we \: have \\ y = {(x - 1)}^{2} \\ by \: matching,\: we \: get \\ 4a = 1 \\ a = \frac{1}{4}$

Now since the parabola is shifted downwards by 4 units, it's focus will be at

$a = \frac{1}{4} - 4 \\ a = - 3 \frac{3}{4}$

So the coordinates of the focal point will be

$f(1, - 3 \frac{3}{4} )$

Now for the directrix,

we know that directrix of a parabola is equidistant from the focus at it's vertex. Hence to find out the directrix line, we find the point on it which is equidistant from the focus.

Distance of focus from vertex

$| - 4 - ( - 3\frac{3}{4} )| \\ | - \frac{1}{4} | \\ \frac{1}{4}$

So the directrix will be below the vertex at a distance of 1/4 units.

$y = - 4 - \frac{1}{4} \\ y = - 4 \frac{1}{4}$

The latus rectum will be the distance between the focus and the corresponding point of the parabola. This implies that the line reflected along the surface of the parabola will be normal to the incident line, i.e., angle of reflection will be a right angle.

To find out the latus rectum, we find the distance between the focus/axis and the x value corresponding to the y value of the focus

We find this by replacing the y value with -15/3 or -3(3/4) and then solve for x.

$y = {(x - 1)}^{2} - 4 \\ \frac{ - 15}{3} = {(x - 1)}^{2} - 4 \\ 4 - \frac{15}{3} = {(x - 1)}^{2} \\ \frac{1}{4} = {(x - 1)}^{2} \\ \frac{1}{2} = x - 1 \\ x = \frac{3}{2}$

Distance of the x coordinates

$d = |1 - \frac{3}{2} | \\ d = \frac{1}{2}$

Hence the latus rectum will be 0.5 units