Conic Sections (Parabola)

Question

1) Find extremities of Latus rectum of parabola $y=x^2-2x+3$.

2) Find the Latus rectum and equation of parabola whose vertex is origin and directrix is $x+y=2$.

Answer 1

Hint: $1) \quad (y-2)=(x-1)^2 $ and in a parabola the lactus retum is $L = 4p$. And the general parabola equation is $(x-h)^2 = 4p(y-k)$

$2) \quad p$ in general parabola equation is equal the distance between vertice and directrix

Answer 2

To calculate extremities of Latus rectum of a parabola we can use the following generalization of the function of parabola: $$ f(x)=\frac{1}{2(b-k)}(a-x)^2+\frac{(b+k)}{2} \tag{1} $$

In which $a$ and $b$ are the coordinates of a focus of a parabola, that is point $(a,b)$. and $k$ is the height of the directrix line from the $x$ axis, that is the function $y=k$. Equation (1) can be easily calculated from the very definition of a focus of a parabola, That the distance from the focus to some point $p$ is equal to the distance from directrix to point $p$, That is: $$y-k=\sqrt{(b-y)^2+(a-x)^2} \tag{2} $$

In which $x$ and $y$ are the coordinates of the point $p$, that's $(x,y)$. We can also derive the function (1) for a general function of a parabola, and we get:

$$ f(x)=\underbrace{\frac{1}{2(b-k)}}_d x^2-\underbrace{\frac{a}{(b-k)}}_hx+\underbrace{\frac{(b^2-k^2+a^2)}{2(b-k)}}_c \tag{3} $$

General function of a parabola:

$$f(x)=dx^2+hx+c \tag{4} $$

In which $d,h$ and $c$ are constants, $a$ and $b$ are coordinates of the focus of a parabola and $k$ is the height of the directrix line from the $x$ axis. In your case $d=1$ , $h=-2$ and $c=3$. Knowing this we can create the following system of equations:

$$ \begin{cases} \frac{1}{2(b-k)}=1 & \\ \frac{a}{(b-k)}=-2 & \\ \frac{(b^2-k^2+a^2)}{2(b-k)}=3 \end{cases} \tag{5} $$

From which we can calculate the values of $k,a$ and $b$ since we have 3 equations and 3 unknowns. To find the extremities of the latus rectum of this parabola we must find zeros of the following equation:

$$\frac{1}{2(b-k)} x^2-\frac{a}{(b-k)}x+\frac{(b^2-k^2+a^2)}{2(b-k)}\textbf{-b}=0 \tag{6} $$

This is a quadratic equation and of course it will have 2 solutions, Which will be $x$ axis coordinates of the points of the extremities of the latus rectum, Namely $x_1$ and $x_2$. Notice that we are subtracting the height of the extremities from the formula of a parabola to get the $x$ axis coordinates of the extremities. Since the latus rectum is at a height of a focus, We can now write the coordinates of the extremities, Which are: $(x_1,b)$ and $(x_2,b)$.

I recommend using Desmos to graph these functions to get a more intuitive idea of this topic.

Answer 3

Answer 1 -

$y = x^2 - 2x + 3$

$x^2 - 2x = y - 3$

$x^2 - 2x + 1 = y - 3 + 1$

$(x - 1)^2 = 1(y - 2)$

let X = x - 1 and Y = y - 2

then $X^2 = Y$

General form of parabola $x^2=4ay$.

On comparing $a = \frac{1}{4}$

Extremites of lactus rectum are (2a, a) and (-2a, a)

So we have extremities (1/2, 1/4) and (-1/2, 1/4).

But these points are with respect to coordinate system having (x,y) as general point.

Therefore (x - 1, y - 2) = (X, Y) = (1/2, 1/4), (-1/2, 1/4)

From here we get (x, y) = ( 1/2, 9/4), (3/2, 9/4)

Answer 2 -

You have vertex and directrix.

Here's link for similar question. I hope you can solve it by yourself taking reference. So I leave this question for you.