1. Parametric equations
Problem solving

2. Parametric equations: problem solving
KUS objectives
BAT solve problems with parametric equations
Starter: write the Cartesian equation of
𝑥=2𝑡 , 𝑦= 𝑡 2
𝑥= 1 𝑡 , 𝑦=2𝑡− 𝑡 2
𝑥=1+ cos 𝑡 , 𝑦= sin 𝑡
Geogebra: parametric eqns

3.  Sub y = 0 in to find t at these points
WB The diagram shows a sketch of the curve with Parametric equations: 𝑥=𝑡− 𝑦=4− 𝑡 2
The curve meets the x-axis at the points A and B. Find their coordinates.
At points A and B, y = 0
 Sub y = 0 in to find t at these points A B
𝑦=4− 𝑡 2
𝑥=𝑡−1
𝑥=𝑡−1
0=4− 𝑡 2
𝑥=(2)−1
𝑥=(−2)−1
𝑡 2 =4
𝑥=1
𝑥=−3
𝑡=±2
 A and B are (-3,0) and (1,0)

4. So the t value at the coordinate (2,0) is -2
WB13 A curve has Parametric equations: 𝑥=𝑎𝑡 𝑦=𝑎( 𝑡 3 +8)
Where a is a constant. Given that the curve passes through (2,0), find the value of a
We know there is a coordinate where x = 2 and y = 0, Sub y = 0 into its equation
𝑦=𝑎( 𝑡 3 +8)
0=𝑎( 𝑡 3 +8)
𝑡 3 +8=0
𝑡 3 =−8
𝑡=−2
So the t value at the coordinate (2,0) is -2
𝑥=𝑎𝑡
(2)=𝑎(−2)
−1=𝑎
Since we know that at (2,0), x = 2 and t = -2, we can put these into the x equation to find a
𝑥=𝑎𝑡
𝑦=𝑎( 𝑡 3 +8)
𝑥=−𝑡
𝑦=−( 𝑡 3 +8)

5. WB14 A curve has Parametric equations: 𝑥= 𝑡 2 𝑦=4𝑡
The line x + y + 4 = 0 meets the curve at point A.
Find the coordinates of A.
𝑥+𝑦+4=0
The first thing we need to do is to find the value of t at coordinate A
 Sub x and y equations into the line equation
( 𝑡 2 )+(4𝑡)+4=0
𝑡 2 +4𝑡+4=0
(𝑡+2) 2 =0
𝑡=−2
So t = -2 where the curve and line meet (point A)
We know an equation for x and one for y, and we now have the t value to put into them…
𝑥= 𝑡 2
𝑦=4𝑡
𝑥= (−2) 2
𝑦=4(−2)
𝑥=4
𝑦=−8
The curve and line meet at (4, -8)

6. You should be able to: BAT solve problems with parametric equations
Write one thing you have learned
Write one thing you need to improve

7. END