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Functions, EXT1 F1 EQ-Bank 3 MC

A curve has the equation  \(\dfrac{(y-2)^2}{9}-\dfrac{(x+1)^2}{4}=1\).

Which of the following expresses the curve in parametric form?

  1. \(x=2 \sec \theta-1, \ y=3 \tan \theta+2\)
  2. \(x=2 \sin \theta-1, \ y=3 \cos \theta+2\)
  3. \(x=2 \tan \theta-1, \ y=3 \sec \theta+2\)
  4. \(x=4 \sec \theta-1, \ y=3 \tan \theta+2\)
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\(\Rightarrow C\)

Show Worked Solution

\(\text{Curve is an ellipse} \ \ \Rightarrow \ \ \text {Eliminate B}\)

\(\text{Trig identity:}\)

\(\sin ^2 \theta+\cos ^2 \theta=1 \ \Rightarrow \ \tan ^2 \theta+1=\sec ^2 \theta \ \ \text{(Divide by \(\cos ^2 \theta\))} \)
 

\(\text{Consider option A:}\)

\(x=2 \sec \theta-1 \ \Rightarrow \ \sec \theta=\dfrac{x+1}{2}\)

\(y=3 \tan \theta+2 \ \Rightarrow \ \tan \theta=\dfrac{y-2}{3}\)

\(\dfrac{(y-2)^2}{9}+1=\dfrac{(x+1)^2}{4} \quad\)X

 
\(\text{Consider option C:}\)

\(x=2 \tan \theta-1 \ \Rightarrow \ \tan \theta=\dfrac{x+1}{2}\)

\(y=3 \sec \theta+2 \ \Rightarrow \ \sec \theta=\dfrac{y-2}{3}\)

\(\dfrac{(x+1)^2}{4}+1=\dfrac{(y-2)^2}{9} \quad \large{\checkmark}\)

\(\Rightarrow C\)

Functions, EXT1 F1 EQ-Bank 4 MC

Determine the Cartesian equation of the circle given by the parametric equations

\(\begin{aligned} & x=-3+4 \cos \theta \\
& y=1+4 \sin \theta\end{aligned}\)

  1. \((x+3)^2+(y-1)^2=4\)
  2. \((x-3)^2+(y+1)^2=4\)
  3. \((x+3)^2+(y-1)^2=16\)
  4. \((x-3)^2+(y+1)^2=16\)
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\(C\)

Show Worked Solution

\(x=-3+4 \cos\, \theta\ \Rightarrow \ \cos\,\theta=\dfrac{x+3}{4} \)

\(y=1+4 \sin\, \theta\ \Rightarrow \ \sin\,\theta=\dfrac{y-1}{4} \)

\(\text{Using}\ \ \sin^{2}\theta + \cos^{2}\theta=1:\)

\( \Big(\dfrac{x+3}{4}\Big)^{2} + \Big(\dfrac{y-1}{4}\Big)^{2}\) \(=1\)  
\((x+3)^2+(y-1)^2\) \(=16\)  

 
\(\Rightarrow C\)

Functions, EXT1 F1 2023 HSC 11a

The parametric equations of a line are given below.

\begin{aligned}
& x=1+3 t \\
& y=4 t
\end{aligned}

Find the Cartesian equation of this line in the form  \(y=m x+c\).  (2 marks)

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\(y=\dfrac{4}{3}x-\dfrac{4}{3} \)

Show Worked Solution

\(x=1+3t\ \ \Rightarrow \ \ t=\dfrac{x-1}{3} \)

\(y\) \(=4t\)  
\(y\) \(=4\bigg{(}\dfrac{x-1}{3}\bigg{)} \)  
\(y\) \(=\dfrac{4}{3}x-\dfrac{4}{3} \)  

Functions, EXT1 F1 2022 HSC 5 MC

A curve is defined in parametric form by  `x=2+t`  and  `y=3-2t^(2)`  for `-1 <= t <= 0`.

Which diagram best represents this curve?
 


 

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`B`

Show Worked Solution

`x=2+t\ \ =>\ \ t=x-2`

`text{Substitute into}\ \ y=3-2t^(2)`

`y=3-2(x-2)^2`

`text{Concave down parabola with maximum at (2, 3)}`

`=>B`

Functions, EXT1 F1 2021 SPEC2 7

A function is defined parametrically by

   `x(t) = 5cos(2t) + 1,\ \ y(t) = 5sin(2t)-1`

If  `A(6, –1)`  and  `B(1, 4)`  are two points that lie on the graph of the function, then find the shortest distance along the graph from `A` to `B`.   (2 marks)

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`(5pi)/2`

Show Worked Solution

`x = 5cos(2t) + 1 \ => \ cos(2t) = (x-1)/5`

`y = 5sin(2t)-1 \ => \ sin(2t) = (y + 1)/5`

`(cos(2t))^2 + (sin(2t))^2` `=1`  
`(x-1)^2 + (y + 1)^2` `=25`  

 

`text(Distance)` `= 1/4 xx 2 xx pi xx r`
  `= (5pi)/2`

Functions, EXT1 F1 2021 HSC 8 MC

The diagram shows a semicircle.
 

Which pair of parametric equations represents the semicircle shown?

  1. `{(x = 3 + sin t),(y = 2 + cos t):} \ \ \ \ \ text(for)\ \ -pi/2 <= t <= pi/2`
  2. `{(x = 3 + cos t),(y = 2 + sin t):} \ \ \ \ \ text(for)\ \ -pi/2 <= t <= pi/2`
  3. `{(x = 3 - sin t),(y = 2 - cos t):} \ \ \ \ \ text(for)\ \ -pi/2 <= t <= pi/2`
  4. `{(x = 3 - cos t),(y = 2 - sin t):} \ \ \ \ \ text(for)\ \ -pi/2 <= t <= pi/2`
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`C`

Show Worked Solution

`text(By elimination:)`

♦♦ Mean mark 31%.

`text(When)\ \ t = pi/2,`

`A(4, 2), \ B(3, 3), \ C(2, 2), \ D(3, 1)`

`->\ text(Eliminate B)`
 

`text(When)\ \ t = -pi/2,`

`A(2, 2), \ C(4, 2), \ D(3, 3)`

`->\ text(Eliminate D)`
 

`text(When)\ \ t = 0,`

`A(3, 3), \ C(3, 1)`

`->\ text(Eliminate A)`
 

`=>\ C`

Functions, EXT1 F1 2019 SPEC2-N 2 MC

The curve given by  `x = 3sec(t) + 1`  and  `y = 2tan(t)-1`  can be expressed in cartesian form as

  1.  `((y + 1)^2)/4-((x-1)^2)/9 = 1`
  2.  `((x + 1)^2)/3-((y-1)^2)/2 = 1`
  3.  `((x-1)^2)/3 + ((y + 1)^2)/2 = 1`
  4.  `((x-1)^2)/9-((y + 1)^2)/4 = 1`
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`D`

Show Worked Solution

`sec^2theta = tan^2theta + 1`

`x = 3sec(t) + 1 \ => \ sec(t) = (x-1)/3`

`y = 2tan(t)-1 \ => \ tan(t) = (y + 1)/2`

`:.((x-1)/3)^2-((y + 1)/2)^2` `= 1`
`((x-1)^2)/9-((y + 1)^2)/4` `= 1`

`=>\ D`

Functions, EXT1 F1 EQ-Bank 11

  1. Find the function described by the following parametric equations

`x = 3t^2` 

`y = 9t, \ \ t > 0`   (1 mark)

  1. Sketch the function.   (1 mark)

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i.    `y = 3sqrt(3x), (t > 0)`

ii.     

Show Worked Solution

i.   `y = 9t \ \ => \ \ t = y/9`

`x` `= 3t^2`
`x` `= 3 · (y/9)^2`
`x` `= y^2/27`
`y^2` `= 27x`
`y` `= 3sqrt(3x), \ \ (t > 0)`

 
ii. 

Functions, EXT1 F1 SM-Bank 9

  1. Sketch the graph of the function described by the parametric equations

     

          `x = 4t - 7`

     

          `y = 2t^2 + t`  (2 marks)

  2. State the domain and range of the function.  (1 mark)

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  1.  
  2. `text(Domain:  all)\ x`
    `text(Range)\ {y: −1/8 <= y < ∞}`
Show Worked Solution

i.   `x = 4t – 7 \ \ => \ \ t = (x + 7)/4`

`y` `= 2t^2 + t`
`y` `= 2((x + 7)/4)^2 + ((x + 7)/4)`
`16y` `= 2(x + 7)^2 + 4(x + 7)`
`16y` `= 2x^2 + 28x + 98 + 4x + 28`
`16y` `= 2x^2 + 32x + 126`
`8y` `= x^2 + 16x + 63`
`y` `= 1/8(x + 7)(x + 9)`

 
`=>\ text(Equation is a concave up quadratic with)`

`text(zeros at)\ \ x = −9\ text(and)\ \ x = −7.`
  

 

ii.   `text(Axis at)\ \ x = −8`

`:.\ y_text(min)` `= 1/8(−1)(1)`
  `= −1/8`

 
`text(Domain: all)\ x`

`text(Range:)\ −1/8 <= y < ∞`

Functions, EXT1 F1 EQ-Bank 10

An equation can be expressed in the parametric form

`x = 2costheta - 1`

`y = 2 + 2sintheta`

  1.  Express the equation in Cartesian form.  (2 marks)

  2.  Sketch the graph.  (1 mark)

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  1. `(x + 1)^2 + (y – 2)^2 = 4`
  2.    
Show Worked Solution
i.    `2costheta` `= x + 1`
  `costheta` `= (x + 1)/2`
`2sintheta` `= y – 2`
`sintheta` `= (y – 2)/2`

 
`text(Using)\ \ cos^2theta + sin^2 = 1:`

`((x + 1)/2)^2 + ((y – 2)/2)^2` `= 1`
`(x + 1)^2 + (y – 2)^2` `= 4`

 

ii.  `text{Sketch circle with centre (−1, 2),  radius = 2}`
 

Functions, EXT1 F1 SM-Bank 8

A circle has the equation  `x^2 - 10x + y^2 + 6y +25 = 0`

  1.  Express the circle in parametric form.  (2 marks)

  2.  Sketch the circle.  (1 mark)

Show Answers Only
  1. `x = 5 + 3costheta`
    `y = −3 + 3sintheta`
  2.   
Show Worked Solution
i.    `x^2 – 10x + y^2 + 6y+25` `= 0`
  `(x – 5)^2 + (y + 3)^2 – 9` `= 0`
  `(x – 5)^2 + (y + 3)^2` `= 9`

 
`=>\ text{Circle centre (5, −3),  radius 3}`
 

`:.\ text(Parametric form is:)`

`x = 5 + 3costheta`

`y = −3 + 3sintheta`

 

ii.  

Functions, EXT1 F1 EQ-Bank 7

The parametric equations of a graph are

`x = 1-1/t`

`y = 1 + 1/t`

Sketch the graph.   (2 marks)

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Show Worked Solution
`x` `= 1-1/t\ \ …\ (1)`
`tx` `= t-1`
`t(1-x)` `= 1`
`t` `= 1/(1-x)\ \ …\ (1^{′})`
`y` `= 1 + 1/t\ \ …\ (2)`

 
`text(Substitute)\ \ t = 1/(1-x)\ \ text{from}\ (1^{′})\ \text{into (2)}:`

`y` `= 1 + 1/(1/(1-x))`
`y` `= 1 + 1-x`
`y` `= 2-x`

 

Functions, EXT1 F1 EQ-Bank 6

The parametric equations of a graph are

`x = t^2`

`y = 1/t`  for  `t > 0`
 

  1. Find the Cartesian equation for the graph.  (1 mark)

  2. Sketch the graph.  (1 mark)

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i.   `y = sqrt(1/x)`

ii.

Show Worked Solution

i.     `x = t^2\ …\ (1)`

`y = 1/t\ …\ (2)`
 

`text(Substitute)\ \ t = 1/y\ \ text{from (1) into (1)}`

`x` `= (1/y)^2`
`y^2` `= 1/x`
`y` `= sqrt(1/x)\ \ \ (y > 0\ \ text(as)\ \ t > 0)`

 

ii.

Functions, EXT1 F1 SM-Bank 4

Sketch the curve described by these two parametric equations

`x = 3cost + 2`

`y = 3sint - 3`   for   `0 <= t < 2pi`.  (3 marks)

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Show Worked Solution

`(x – 2) = 3cost`

`(y + 3) = 3sint`

`(x – 2)^2 + (y + 3)^2` `= (3cost)^2 + (3sint)^2`
  `= 9cos^2t + 9sin^2t`
  `= 9(cos^2t + sin^2t)`
  `= 9`

 
`text(Sketch:) \ (x – 2)^2 + (y + 3)^2 = 3^2`