Question

Consider the Line L(t) = <2+t,5-4t>. Then L intersects:

The x- axis at the point (3.25,0) when t = 5/4

The y- axis at the point (0, 13) when t = -2

The parabola y = x^2 at the points ___ and ___ when t = ____ and ____

Answer 1

The parabola
`y=x^(2)` at the points
`(-2+√(17),21-4√(17))` and
`(-2-√(17),21+4√(17))` when
`t=t_(1)=-4+√(17)` and
`t=t_(2)=-4-√(17)`

Explanation:

We have the following line written in parametric form :

`L(t)=(2+t,5-4t)` with
`t` ∈ IR.

In order to find the intersection between
`L(t)` and the parabola
`y=x^(2)` we know that ''
`2+t`'' is the x-coordinate of the line and ''
`5-4t`'' is the y-coordinate of the line. Now, to solve this problem we need to find the values of ''
`t`'' in which the intersection occurs. We can do this by replacing the components ''
`x`'' and ''
`y`'' of
`L(t)` in the equation of the parabola ⇒

`L(t)=(2+t,5-4t)` = ( x component , y component ) = ( x , y ) ⇒

In the parabola :
`y=x^(2)` ⇒
`5-4t=(2+t)^(2)`

Solving the equation we find that :

`t^(2)+8t-1=0`

Using the quadratic formula with

`a=1` ,
`b=8` and
`c=-1`

We find that the two possible values for t :

`t_(1)=\frac{-b+\sqrt{b^(2)-4ac}}{2a}` and
`t_(2)=\frac{-b-\sqrt{b^(2)-4ac}}{2a}`

are
`t_(1)=-4+√(17)` and
`t_(2)=-4-√(17)`

This values
`t_(1)` and
`t_(2)` are the values of the parameter t where the line intersects the parabola so we can find the points by replacing the values of the parameter in the equation
`L(t)` :

`L(t_(1))=(-2+√(17),21-4√(17))` and

`L(t_(2))=(-2-√(17),21+4√(17))`

The final answer is

The parabola
`y=x^(2)` at the points
`(-2+√(17),21-4√(17))` and
`(-2-√(17),21+4√(17))` when
`t=t_(1)=-4+√(17)` and
`t=t_(2)=-4-√(17)`