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Question: How do you transform parametric equations into cartesian form: \( x = 3 + 2\cos t \) and \[y = 1 + 5...

How do you transform parametric equations into cartesian form: x=3+2cos⁡tx = 3 + 2\cos t and y=1+5sin⁡ty = 1 + 5\sin t

Explanation

Solution

Hint : In order to determine the cartesian equation from the given parametric equation, rewrite both equations such that trigonometric function is left on either side. Now square and add both the equations to obtain an equation. Use the property of trigonometry and identity of (A−B)2=A2+B2−2AB{\left( {A - B} \right)^2} = {A^2} + {B^2} - 2AB and simplify the equation by combining all the like terms to obtain the required result.
Formula:
(A−B)2=A2+B2−2AB{\left( {A - B} \right)^2} = {A^2} + {B^2} - 2AB
sin⁡2x+cos⁡2x=1{\sin ^2}x + {\cos ^2}x = 1

Complete step-by-step answer :
We are given two parametric equation as x=3+2cos⁡tx = 3 + 2\cos t and y=1+5sin⁡ty = 1 + 5\sin t
As per the question, we have to transform the above parametric equations into cartesian equations.
So, the cartesian equation of the curve having parametric equation x=3+2cos⁡tx = 3 + 2\cos t and y=1+5sin⁡ty = 1 + 5\sin t can be obtained by squaring and adding the given parametric equations.
Before squaring , we will be rewriting both the equations, such that only the trigonometric function is left on one side by applying some transformations. We get

x=3+2cos⁡t x−3=2cos⁡t ⇒x−32=cos⁡t   x = 3 + 2\cos t \\\ x - 3 = 2\cos t \\\ \Rightarrow \dfrac{{x - 3}}{2} = \cos t \;

Similarly ,

y=1+5sin⁡t y−1=5sin⁡t ⇒y−15=sin⁡t   y = 1 + 5\sin t \\\ y - 1 = 5\sin t \\\ \Rightarrow \dfrac{{y - 1}}{5} = \sin t \;

Thus, we have parametric equation as x−32=cos⁡t\dfrac{{x - 3}}{2} = \cos t and y−15=sin⁡t\dfrac{{y - 1}}{5} = \sin t .
Now squaring squaring both sides of the equation of both the parametric equation and adding them , we get
(x−32)2+(y−15)2=(sin⁡t)2+(cos⁡t)2{\left( {\dfrac{{x - 3}}{2}} \right)^2} + {\left( {\dfrac{{y - 1}}{5}} \right)^2} = {\left( {\sin t} \right)^2} + {\left( {\cos t} \right)^2}
(x−3)2(2)2+(y−1)2(5)2=sin⁡2t+cos⁡2t\dfrac{{{{\left( {x - 3} \right)}^2}}}{{{{\left( 2 \right)}^2}}} + \dfrac{{{{\left( {y - 1} \right)}^2}}}{{{{\left( 5 \right)}^2}}} = {\sin ^2}t + {\cos ^2}t
As we know the trigonometry identity sin⁡2x+cos⁡2x=1{\sin ^2}x + {\cos ^2}x = 1 , we can rewrite the equation as
(x−3)24+(y−1)225=1\dfrac{{{{\left( {x - 3} \right)}^2}}}{4} + \dfrac{{{{\left( {y - 1} \right)}^2}}}{{25}} = 1
Simplifying the square terms using the identity (A−B)2=A2+B2−2AB{\left( {A - B} \right)^2} = {A^2} + {B^2} - 2AB ,we get
25(x2−6x+9)+4(y2−2y+1)100=1\dfrac{{25\left( {{x^2} - 6x + 9} \right) + 4\left( {{y^2} - 2y + 1} \right)}}{{100}} = 1
Multiplying both sides with 100100 and expanding the terms ,we have

100×25(x2−6x+9)+4(y2−2y+1)100=100×1 25(x2−6x+9)+4(y2−2y+1)=100 25x2−150x+225+4y2−8y+4=100  100 \times \dfrac{{25\left( {{x^2} - 6x + 9} \right) + 4\left( {{y^2} - 2y + 1} \right)}}{{100}} = 100 \times 1 \\\ 25\left( {{x^2} - 6x + 9} \right) + 4\left( {{y^2} - 2y + 1} \right) = 100 \\\ 25{x^2} - 150x + 225 + 4{y^2} - 8y + 4 = 100 \\\

Combining all the like terms,

25x2+4y2−150x−8y+225+4−100=0 25x2+4y2−150x−8y+129=0   25{x^2} + 4{y^2} - 150x - 8y + 225 + 4 - 100 = 0 \\\ 25{x^2} + 4{y^2} - 150x - 8y + 129 = 0 \;

∴25x2+4y2−150x−8y+129=0\therefore 25{x^2} + 4{y^2} - 150x - 8y + 129 = 0
Therefore, the required cartesian equation is 25x2+4y2−150x−8y+129=025{x^2} + 4{y^2} - 150x - 8y + 129 = 0 .
So, the correct answer is “25x2+4y2−150x−8y+129=025{x^2} + 4{y^2} - 150x - 8y + 129 = 0”.

Note : 1.To convert the given parametric equation to the cartesian equation, we have to first eliminate the variable tt from both the given equation. To do so use , square and add the parametric equations.
2. Avoid any jump of step in the solution in such problems, as it might increase the change of error in the end result.
3. While expanding the terms, remember to write terms with proper signs.
4. Like terms are the terms having the same variable with the same degree.