ANSWER
[tex]y = 2x -4[/tex]
EXPLANATION
Part a)
Eliminating the parameter:
The parametric equation is
[tex]x = 5 + ln(t) [/tex]
[tex]y = {t}^{2} + 5[/tex]
From the first equation we make t the subject to get;
[tex]x - 5 = ln(t) [/tex]
[tex]t = {e}^{x - 5} [/tex]
We put it into the second equation.
[tex]y = { ({e}^{x - 5}) }^{2} + 5[/tex]
[tex]y = { ({e}^{2(x - 5)}) } + 5[/tex]
We differentiate to get;
[tex] \frac{dy}{dx} = 2 {e}^{2(x - 5)} [/tex]
At x=5,
[tex] \frac{dy}{dx} = 2 {e}^{2(5 - 5)} [/tex]
[tex]\frac{dy}{dx} = 2 {e}^{0} = 2[/tex]
The slope of the tangent is 2.
The equation of the tangent through
(5,6) is given by
[tex]y-y_1=m(x-x_1)[/tex]
[tex]y - 6 = 2(x - 5)[/tex]
[tex]y = 2x - 10 + 6[/tex]
[tex]y = 2x -4[/tex]
Without eliminating the parameter,
[tex] \frac{dy}{dx} = \frac{ \frac{dy}{dt} }{ \frac{dx}{dt} } [/tex]
[tex]\frac{dy}{dx} = \frac{ 2t}{ \frac{1}{t} } [/tex]
[tex]\frac{dy}{dx} = 2 {t}^{2} [/tex]
At x=5,
[tex]5 = 5 + ln(t) [/tex]
[tex] ln(t) = 0[/tex]
[tex]t = {e}^{0} = 1[/tex]
This implies that,
[tex]\frac{dy}{dx} = 2 {(1)}^{2} = 2[/tex]
The slope of the tangent is 2.
The equation of the tangent through
(5,6) is given by
[tex]y-y_1=m(x-x_1)[/tex]
[tex]y - 6 = 2(x - 5) =[/tex]
[tex]y = 2x -4[/tex]
