Question

Find an equation of the tangent to the curve at the given point by both eliminating the parameter and without eliminating the parameter.x = 6 + ln(t), y = t2 + 6, (6, 7)

Answer 1

Answer:2x-y=5

Step-by-step explanation:

Given

$x=6+\ln t$

$y=t^{2}+6$

$\left ( a\right )$ without eliminating parameter

$\frac{\mathrm{d} x}{\mathrm{d} t}$=$\frac{1}{t}$

$\frac{\mathrm{d} y}{\mathrm{d} t}$=2t

$\frac{\mathrm{d} y}{\mathrm{d} x}=2t^2$

at $\left ( 6,7\right )$

6=6+$\ln\left ( t\right )$

t=1

Equation of line is given by

2=$\frac{y-7}{x-6}$

2x-12=y-7

2x-y=5

$\left ( b\right )$by eliminating parameter

$x-6=\ln \left ( t\right )$

$t=e^{x-6}$

$y=t^2 +6$

$y=e^{\left ( 2x-12\right )}+6$

differentiating we get

$\frac{\mathrm{d}y}{\mathrm{d} x}=2e^\left ( 2x-12\right )$

$at \left ( 6,7\right )$

$\frac{\mathrm{d}y}{\mathrm{d} x}=2$

$2\left ( x-6\right )=\left ( y-7\right )$

2x-y=5