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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ch4aika [34]

Answer:

The number that belongs in the green box is equal to 909.

General Formulas and Concepts:
Algebra I

Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtraction Property of Equality

Trigonometry

[Right Triangles Only] Pythagorean Theorem:
$\displaystyle a^2 + b^2 = c^2$

a is a leg
b is another leg
c is the hypotenuse

Step-by-step explanation:

Step 1: Define

Identify given variables.

a = 30

b = 3

c = x

Step 2: Find x

Let's solve for the general equation that allows us to find the hypotenuse:

[Pythagorean Theorem] Square root both sides [Equality Property]:
$\displaystyle \begin{aligned}a^2 + b^2 = c^2 \rightarrow c = \sqrt{a^2 + b^2}\end{aligned}$

Now that we have the formula to solve for the hypotenuse, let's figure out what x is equal to:

[Equation] Substitute in variables:
$\displaystyle \begin{aligned}c & = \sqrt{a^2 + b^2} \\x & = \sqrt{30^2 + 3^2}\end{aligned}$
Evaluate:
$\displaystyle \begin{aligned}c & = \sqrt{a^2 + b^2} \\x & = \sqrt{30^2 + 3^2} \\& = \boxed{ \sqrt{909} } \\\end{aligned}$