Eliminate the parameter for the following set of parametric equations: x= t^2 + 2 y= 4t^2

Mathematics

1 answer:

mr_godi [17]3 years ago

6 0

Answer:

Solution : y = 4x - 8

Step-by-step explanation:

The first thing we want to do is isolate t², rather than t. Why? As you can see when we substitute t² into the second equation, it will be easier than substituting t, as t is present in the form t². So, let's isolate t² in the first equation --- ( 1 )

x = t² + 2,

t² = x - 2

Now let's substitute this value of t² in the second equation --- ( 2 )

y = 4t²,

y = 4(x - 2),

y = 4x - 8 ~ And hence our solution is option c.

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A building has an entry the shape of a parabolic arch 84 ft high and 42 ft wide at the base, as shown below. A parabola opening

lara [203]

Answer:

Step-by-step explanation:

Because of the nature of the information we are given, we have no choice but to use the equation

$y=a(x-h)^2+k$

and solve for a.

We know by the info that the vertex is (0, 84). We also know that if the vertex is at the origin, and that the base is 42 feet wide, it spans 21 feet to the right of the origin and 21 feet to the left of the origin. That means that we have 2 coordinates from which we need to pick one for our x and y in the equation. I don't like negatives, so I am going to choose the coordinate (21, 0) as x and y. Because this parabola opens upside down, as archways of door openings do, our "a" value better come out algebraically as a negative. Let's see...From the vertex we have that h = 0 and k = 84. So filling in:

$0=a(21-0)^2+84$ and simplifying a bit: