Answer:
[tex]y=2x-8[/tex]
Step-by-step explanation:
The given parametric equation is;
[tex]x=6+ln(t),y=t^2+3[/tex]
To eliminate the parameter we make [tex]t[/tex] the subject in one equation and put it inside the other.
We make [tex]t[/tex] the subject in [tex]x=6+ln(t)[/tex] because it is easier.
[tex]\Rightarrow x-6=ln(t)[/tex]
[tex]\Rightarrow {e}^{x-6}=e^{ln(t)}[/tex]
[tex]\Rightarrow {e}^{x-6}=t[/tex]
Or
[tex]t={e}^{x-6}[/tex]
We now substitute this into [tex]y=t^2+3[/tex].
This gives us;
[tex]y=(e^{x-6})^2+3[/tex].
[tex]\Rightarrow y=e^{2(x-6)}+3[/tex].
We have now eliminated the parameter.
The equation of the tangent at (6,4) is given by;
[tex]y-y_1=m(x-x_1)[/tex]
where the gradient function is given by;
[tex]\frac{dy}{dx}=2e^{2(x-6)}[/tex]
We substitute [tex]x=6[/tex] into the gradient function to obtain the gradient.
[tex]\Rightarrow m=2e^{2(6-6)}[/tex]
[tex]\Rightarrow m=2e^0[/tex]
[tex]\Rightarrow m=2[/tex]
The equation of the tangent becomes
[tex]y-4=2(x-6)[/tex]
We simplify to obtain
[tex]y=2x-12+4[/tex]
[tex]y=2x-8[/tex]
The given parametric equation is;
[tex]x=6+ln(t),y=t^2+3[/tex]
For [tex]x=6+ln(t)[/tex]
[tex]\frac{dx}{dt}=\frac{1}{t}[/tex]
For [tex]y=t^2+3[/tex]
[tex]\frac{dy}{dt}=2t[/tex]
The slope is given by;
[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}=2t^2[/tex]
At the point, (6,4), we plug in any of the values into the parametric equation and find the corresponding value for [tex]t[/tex].
Notice that
When [tex]x=6[/tex], [tex]6=6+\ln(t)[/tex]
[tex]6-6=\ln(t)[/tex]
[tex]0=\ln(t)[/tex]
[tex]e^0=e^\ln(t)[/tex]
[tex]1=t[/tex]
when [tex]y=4[/tex], [tex]4=t^2+3[/tex]
[tex]4-3=t^2[/tex]
[tex]1=t^2[/tex]
[tex]t=\pm1[/tex]
But the slope is the same when we plug in any of these values for t.
[tex]\frac{dy}{dx}=2(\pm1)^2=2[/tex]
The equation of the tangent becomes
[tex]y-4=2(x-6)[/tex]
We simplify to obtain
[tex]y=2x-12+4[/tex]
[tex]y=2x-8[/tex]
