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Chapter 10: Q. 37 (page 695)

In Problems 35–38, find parametric equations that define the curve shown.

Short Answer

Expert verified

The parametric equation is $x = 3 \cos t, y = 2 \sin t, 0 \leq t \leq 2 π$

Step by step solution

01

Step 1. Given Information

We are given a graph

Given the curve and its coordinates, we were asked to determine its parametric equation. To do that, we have to find equation representing the given graph to find the parametric equation.

02

Step 2. Finding Parametric Equation

Since the ellipse is centre on origin, we can use the equation

$x = a \cos t, y = b \sin t$

We have $a = 3, b = 2$

So, role="math" localid="1648033258757" $x = 3 \cos t, y = 2 \sin t, 0 \leq t \leq 2 π$

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Most popular questions from this chapter

In Problems 31– 42, rotate the axes so that the new equation contains no xy-term. Analyze and graph the new equation. Refer to Problems 21–30 for Problems 31– 40. $11 x^{2} + 10 \sqrt{3} x y + y^{2} - 4 = 0$

In Problem, identify the graph of each equation without completing the squares.

$x^{2} + 4 x + y + 3 = 0$

To graph $y = {(x - 3)}^{2} + 1$ , shift the graph of $y = x^{2}$ to the right________ units and then________ 1 unit.

Find the equation of the parabola described. Find the two points that define the latus rectum, and graph the equation by hand.

Vertex is at $(4, - 2)$ and focus is at $(6, - 2)$ .

In Problems 15–18, the graph of a hyperbola is given. Match each graph to its equation.

Equations:

$(A) \frac{x^{2}}{4} - y^{2} = 1 (B) x^{2} - \frac{y^{2}}{4} = 1 (C) \frac{y^{2}}{4} - x^{2} = 1 (D) y^{2} - \frac{x^{2}}{4} = 1$

Graph:

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