Question

1) Consider the Cartesian equation of a hyperbola, y=±4√1+x^2/9, versus the parametric equations of the same hyperbola:

Answer 1

Answer:

Step-by-step explanation:

1) Consider the cartesian equation of a hyperbola

$y= \pm4\sqrt{1+\frac{x^2}{9} }$

whose parametric equation are x = 3tan t and y = 4sec t

The parametric equation show the direction in which the parabola is drawn.

For a given value of the independent variable, the parametric equation yield exactly one point on the graph and

with parametric equations, each arm of the hyperbola is graphed in a continuous manner on the graphing calculator.

With parametric equation, the plotting becomes easier.

The cartesian equation of the hyperbola is a quadratic equation. So for a single x( or single y), there will be two values for y(or x). In parametric form, for a single 't', there will be only one point corresponding to that t.

2)

$If \left \{ {{x=3t} \atop {y= t^{2} }} \right. then t= \frac{x}{3} and y=( \frac{t}{3} )^2= \frac{t^{2}}{9} .$

You can match points:

if x=0, y=0

if x=3, y=1

if x=-3, y=1

if x=6, y=4

if x=-6, y=4

The graph is in attached file

3)

Given that a conic section has  parametric equations x= a cos t and y= b sin t,

Since only sum of squares of sin and cos =1 we find that out of conic sections, namely

parabola, circle, ellipse, hyperbola this can correspond only to ellipse or circle

Because parabola has only one variable in 2 degrees and hyperbola is difference of squares.

i) When a=b, we have this represents a  circle with radius a.

ii) When a>b, we get an ellipse horizontal with major axis horizontal and centre at the origin and vertex at (a,0) (-a,0) (0,b) (0,-b)

iii) When a <b, we get a vertical ellipse with major axis as y axis and vertices same as

(a,0) (-a,0) (0,b) (0,-b)

4)

the main equation parametric of an  ellipse is 

x²/a² + y²/b² = 1

a≠0 and b≠0

let's consider x=3 cos t and y=8 sin t, these are equivalent to x²=9 cos²t and y²=64 sin²t, and imiplying  x²/9=cos²t and y²/64=sin²t

therefore, x²/9+y²/64= cos²t+ sin²t, but we know that cos²t+ sin²t =1 (trigonometric fundamental rule)

so finally,  x²/9+y²/64=1 equivalent of  x²/3²+y²/8²= 1

this is an ellipse

with the same method, we found

x²/9+y²/64= cos²4t+ sin²4t =1, so the only difference between the graphs is the value of the angle (t and 4t)