Find an equation of the tangent line at a point
$( 0,0 ) \text { on the curve }$ $x = t ^ { 2 } - 4 , y = t ^ { 2 } - 2 t$
A) $2 y + x = 0$
B) $y - x = 0$
C) $y - 2 x = 0$
D) $y + 2 x = 0$
E) $2 y - x = 0$

Question

Find an equation of the tangent line at a point 
$( 0,0 ) \text { on the curve }$ $x = t ^ { 2 } - 4 , y = t ^ { 2 } - 2 t$
A) $2 y + x = 0$
B) $y - x = 0$
C) $y - 2 x = 0$
D) $y + 2 x = 0$
E) $2 y - x = 0$

Quizplus · Accepted Answer

To find the equation of the tangent line at a point, we need to find the derivative of the curve at that point and then use the point-slope form of a line. 
First, we find the derivative of the curve: 
$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{(2t-2)}{(2t)} = \frac{t-1}{t}$
Next, we find the slope of the tangent line at point (0,0) by plugging in $t=0$: 
$\frac{dy}{dx}\bigg|_{(0,0)} = \frac{0-1}{0} = 	ext{undefined}$
This means the tangent line is vertical and has the equation $x=0$. 
However, none of the answer choices are in the form $x=$, so we need to manipulate the equation. 
The given curve is parameterized by $t$, so we can re-write $x=t^2-4$ as $t=\pm\sqrt{x+4}$. Substituting this into $y=t^2-2t$ gives $y=(x+4)-2\sqrt{x+4}$ or $y=(x+4)+2\sqrt{x+4}$. 
Taking the derivative of either of these gives: 
$\frac{dy}{dx} = 1- \frac{1}{\sqrt{x+4}} = \frac{\sqrt{x+4}-1}{\sqrt{x+4}}$
Again, we plug in $x=0$ to get the slope of the tangent line at point (0,0):
$\frac{dy}{dx}\bigg|_{(0,0)} = \frac{\sqrt{4}-1}{\sqrt{4}} = \frac{1}{2}$
Finally, we use the point-slope form to get the equation of the tangent line: 
$y-0 = \frac{1}{2}(x-0)$
Simplifying gives $2y-x=0$, which matches answer choice E.

Find an Equation of the Tangent Line at a Point (0,0) on the curve ( 0,0 ) \text { on the curve }(0,0) on the curve

Find an Equation of the Tangent Line at a Point $( 0,0 ) \text { on the curve }$