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

Answer:

The parabola $y=x^{2}$ at the points $(-2+\sqrt{17},21-4\sqrt{17})$ and $(-2-\sqrt{17},21+4\sqrt{17})$ when $t=t_{1}=-4+\sqrt{17}$ and $t=t_{2}=-4-\sqrt{17}$

Step-by-step 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 $t_{1}=-4+\sqrt{17}$ and $t_{2}=-4-\sqrt{17}$

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 $L(t)$ :

$L(t_{1})=(-2+\sqrt{17},21-4\sqrt{17})$ and

$L(t_{2})=(-2-\sqrt{17},21+4\sqrt{17})$

The final answer is

The parabola $y=x^{2}$ at the points $(-2+\sqrt{17},21-4\sqrt{17})$ and $(-2-\sqrt{17},21+4\sqrt{17})$ when $t=t_{1}=-4+\sqrt{17}$ and $t=t_{2}=-4-\sqrt{17}$