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Dominik [7]
3 years ago
10

The following two sets of parametric functions both represent the same ellipse. Explain the difference between the graphs.

Mathematics
2 answers:
statuscvo [17]3 years ago
7 0
The answer
ellipse main equatin is as follow:

X²/ a²   +  Y²/ b²  =1, where a≠0 and b≠0

for the first equation: <span>x = 3 cos t and y = 8 sin t
</span>we can write <span>x² = 3² cos² t and y² = 8² sin² t
and then  </span>x² /3²= cos² t and y²/8² =  sin² t
therefore,  x² /3²+ y²/8²  =  cos² t + sin² t = 1
equivalent to x² /3²+ y²/8²  = 1

for the second equation, <span>x = 3 cos 4t and y = 8 sin 4t we found
</span>x² /3²+ y²/8²  = cos² 4t + sin² 4t=1

ehidna [41]3 years ago
7 0

Answer with explanation:

The two parametric equation of the same ellipse is

x = 3 cos t and y = 8 sin t

x = 3 cos 4 t and y = 8 sin 4 t

\frac{x}{3}=\cos t\\\\ \frac{x}{3}=\cos 4t\\\\ \frac{y}{8}=\sin t\\\\ \frac{y}{8}=\sin 4t\\\\ (\frac{x}{3})^2+ (\frac{y}{8})^2=\sin^2t+\cos^2t \text{or}=\sin^2 4t+\cos^2 4t=1\\\\\frac{x^2}{9}+\frac{y^2}{64}=1

This is the equation of same ellipse, having different Parametric forms.

→The function involving , 3 cos t and 8 sin t has maximum value, 3 and 8,respectively , and a period of π  , whereas, the function  3 cos 4 t and , 8 sin 4 t , has also same maximum value, 3 and 8,respectively , but period changes , the period after which cycle of trigonometric function  sin 4 t and cos 4 t  repeats is, t=\frac{\pi}{4}.

→x = 3 cos t and y = 8 sin t

\frac{x}{y}=\frac{3}{8 \tan t}\\\\y=\frac{8 \tan t*x}{3}

→x = 3 cos 4 t and y = 8 sin 4 t

\frac{x}{y}=\frac{3}{8 \tan 4 t}\\\\y=\frac{8 \tan 4 t*x}{3}

Also, these are equation of two lines having different slopes both passing through the origin.

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General Idea:

When we are given a point P(x, y) centered at origin with a scale factor of k, then the dilated point will be given by P' (kx, ky)

Applying the concept:

In the diagram given, the coordinate of C is (6, 3). Triangle ABC is being dilated with the center of dilation at the origin. The image of C, point C' has coordinates of (7.2, 3.6).

If C (x, y) centered at origin with a scale factor of k, then the dilated point will be given by C' (kx, ky)

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Substituting 6 for x in the equation kx = 7.2, we get 6k = 7.2.

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Answer:

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