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Sagot :
Answer:
1) y = 2x + 1
2) y = 2x + 1
Step-by-step explanation:
The parametric equation given is;
x = 1 + ln t and y = t² + 2 at (1, 3)
1) without eliminating the parameter;
Using, x = 1 + ln t ;
dx/dt = 1/t
Using y = t² + 2;
dy/dt = 2t
Slope which is dy/dx is gotten from;
dy/dx = (dy/dt)/(dx/dt)
dy/dx = 2t/(1/t)
dy/dx = 2t²
For x = 1 + In t, at x = 1, we have;
1 = 1 + In t
In t = 0
t = 1
For y = t² + 2, at y = 3, we have;
3 = t² + 2
t² = 3 - 2
t² = 1
t = ±1
Since t = ±1, then;
dy/dx = 2(±1)²
dy/dx = 2
Equation of the tangent is;
y - 3 = 2(x - 1)
y - 3 = 2x - 2
y = 2x - 2 + 3
y = 2x + 1
2) By eliminating the parameter
x = 1 + In t
Let's make t the subject of the equation.
In t = x - 1
t = e^(x - 1)
Let's put e^(x - 1) for t in y = t² + 2
Thus;
y = e^(x - 1)² + 2
y = e^(2(x - 1)) + 2
Thus, parameter has been eliminated
Equation of the tangent is gotten from;
y - y1 = m(x - x1)
m is gradient = dy/dx = 2e^(2(x - 1))
at (1, 3), we have x = 1. Thus;
m = 2e^(2(1 - 1))
m = 2e^0
m = 2
Thus, equation of tangent at (1,3) is;
y - 3 = 2(x - 1)
y - 3 = 2x - 2
y = 2x - 2 + 3
y = 2x + 1
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