The curve $x = 2 - 4 \cos ^ { 2 } t , \quad y = \tan t \left( 1 - 2 \cos ^ { 2 } t \right)$ cross itself at some point $\left( x _ { 0 } , y _ { 0 } \right)$ . Find the equations of both tangent lines at that point.
A) $y = \frac { x } { 2 } - 4 , y = - \frac { x } { 2 } - 2$
B) $y = \frac { x } { 2 } , y = - \frac { x } { 2 }$
C) $y = \frac { x } { 2 } , y = - \frac { x } { 4 }$
D) $y = \frac { x } { 2 } + 8 , y = - \frac { x } { 2 } + 2$
E) $y = x + 2 , y = - x + 2$

Question

Quizplus · Accepted Answer

To find the point of self-intersection, we need to solve $x=2-4\cos^2 t$ and $y=	an t\left(1-2\cos^2 tight)$ simultaneously. From the equation for $x$, we have $\cos^2 t=\frac{2-x}{4}$, and substituting this into the equation for $y$, we get $y=	an t\left(\frac{x-2}{2}ight)$. Since $	an t$ is not zero at a point of self-intersection, we must have $\frac{x-2}{2}=0$, which gives $x=2$. To find the corresponding $y$-coordinate, we substitute $x=2$ into the equation for $y$ and get $y=	an t\left(-\frac{1}{2}ight)$. This means that $	an t=-2y$, so $\cos^2 t=\frac{2-x}{4}=0$, which in turn means that $\sin^2 t=1$, i.e. $\sin t=\pm 1$. Thus, the two points of intersection are $(2,-2)$ and $(2,2)$.

To find the tangent lines at $(2,-2)$, we need to find $\frac{dy}{dx}$ at this point. From the equation for $y$, we have $y=	an t\left(1-2\cos^2 tight)$, so $\frac{dy}{dt}=\frac{d}{dt}\left(	an t\left(1-2\cos^2 tight)ight)=\sec^2 t\left(1-4\cos t\sin tight)$. Since $\cos t=\frac{2-x}{4}=0$ and $\sin t=-\frac{1}{2}$ at $(2,-2)$, we have $\frac{dy}{dt}=\sec^2 t=4$. Also, since $\frac{dx}{dt}=8\sin t\cos t$, we have $\frac{dx}{dy}=\frac{dx/dt}{dy/dt}=\frac{2\sin t}{\sec^2 t}=2\sin t\cos t=-1$. Thus, the equation of the tangent line at $(2,-2)$ is $y-(-2)=-1(x-2)$, i.e. $y=-x+4$.

The same procedure can be used to find the tangent line at $(2,2)$, or we can use symmetry. Since the curve is symmetric about the $x$-axis, the tangent line at $(2,-2)$ is parallel to the tangent line at $(2,2)$ (and equidistant from it, since the curve is also symmetric about the line $y=0$). Thus, the equation of the tangent line at $(2,2)$ is also $y=-x+4$. Therefore, the answer is $\boxed{	extbf{(B)}}$.

The Curve x=2−4cos⁡2t,y=tan⁡t(1−2cos⁡2t)x = 2 - 4 \cos ^ { 2 } t , \quad y = \tan t \left( 1 - 2 \cos ^ { 2 } t \right)x=2−4cos2t,y=tant(1−2cos2t)

The Curve $x = 2 - 4 \cos ^ { 2 } t , \quad y = \tan t \left( 1 - 2 \cos ^ { 2 } t \right)$