There are two tangent lines from the point (0 1)

The user kccu already found your mistake. Here is a shorter way to do the question.

The curve is $(5\cos t, 3\sin t\,\cos t)$. With $0\leq t<2\pi$, it goes through the origin two times: when $t=\pi/2$ and when $t=3\pi/2$. Then derivative is, as you say, $$ (-5\sin t, 3\cos 2t). $$ At the two values of $t=\pi/2$ and $t=3\pi/2$ we get $$ (-5,-3),\ \ \text{ and }\ \ (5,-3). $$ The lines are then $(x,y)=t(-5,-3)$ and $(x,y)=t(5,-3)$. That is $$ y=-\frac35\,x \ \ \ \text{ and } y=\frac35\,x. $$ So the slopes are $3/5$ and $-3/5$.

The question: There are two tangent lines from the point (0,1) to the circle x^2 + (y+1)^2=1 [centered at (0,-1) with a radius of one]. Find equations of these lines by using the fact that each tangent line intersects the circle in exactly one point.

There is also a graph but it shows pretty much what the question says. I have no idea how I can find the slopes of these tangent lines

If we solve the circle equation for y, we get $\displaystyle y=\sqrt{1-x^{2}}-1$ We can use $\displaystyle y-y_{1}=m(x-x_{1})$, plug in all of our info wrt x and solve for x. Where m is the derivative of y, $\displaystyle y_{1}=1, \;\ x_{1}=0$ $\displaystyle \underbrace{\sqrt{1-x^{2}}-1}_{\text{y}}-1=\frac{-x}{\sqrt{1-x^{2}}}(x-0)$ Now, solve for x. There will be the two values of x. This will be the two x values where the lines are tangent to the circle.

Then, you will have all the info needed to find the line equations.

Last edited: Dec 17, 2008

Reactions: atruenut

Ah thank you very much! (Clapping)
But I'm wondering if there is a way to find the derivative without using calculus since this problem was in the review section of the calc book?

Ah thank you very much! (Clapping)
But I'm wondering if there is a way to find the derivative without using calculus since this problem was in the review section of the calc book?

From the equation of the circle you know that the center of the circle is: C(0, -1). Let T(x, y) denote the tangent point on the circle then the slope of $\displaystyle \overline{TC}$ is: $\displaystyle m_{TC}=\dfrac{y-x_C}{x-x_C}=\dfrac{y-(-1)}{x-0}$ Since T is placed on the circle line the coordinates of T are $\displaystyle T(x, \sqrt{1-x^2} - 1)$ Therefore $\displaystyle m_{TC}=\dfrac{\sqrt{1-x^2} - 1-(-1)}{x}=\dfrac{\sqrt{1-x^2}} x$ The line TC is perpendicular to the tangent because TC = r. Therefore the slope of the tangent must be the negative reciprocal of the slope of TC:

$\displaystyle m_t=-\dfrac x{\sqrt{1-x^2}}$

Analytic approach

The question: There are two tangent lines from the point (0,1) to the circle x^2 + (y+1)^2=1 [centered at (0,-1) with a radius of one]. Find equations of these lines by using the fact that each tangent line intersects the circle in exactly one point.
There is also a graph but it shows pretty much what the question says. I have no idea how I can find the slopes of these tangent lines

Let $\displaystyle C:~x^{2}+(y+1)^{2}=1$. Clearly, the center of the circle is $\displaystyle M=(0,-1)$ and the radius is $\displaystyle r=1$. Let $\displaystyle Q:=(q_{1},q_{2})$ be an arbitrary point on $\displaystyle C$. Then, it is easy to see that the tangent line of the circle at $\displaystyle Q$ is given by $\displaystyle \langle\underset{\overrightarrow{QX}}{\underbrace{(x-q_{1},y-q_{2})}},\underset{\overrightarrow{MQ}}{\underbrace{(q_{1},q_{2}+1)}}\rangle=0$ or simply

$\displaystyle q_{1}(x-q_{1})+(y-q_{2})(q_{2}+1)=0$.......................(*)

We want this line pass through the point $\displaystyle (0,1)$. Then, we have $\displaystyle 0=q_{1}^{2}+(q_{2}-1)(q_{2}+1)=q_{1}^{2}+q_{2}^{2}-1$, using the fact that $\displaystyle Q\in C$, we have $\displaystyle q_{1}^{2}+q_{2}^{2}=-2q_{2}$, therefore, we obtain $\displaystyle q_{2}=-1/2$. Using this in the circle's equation, we have $\displaystyle q_{1}=\pm\sqrt{3}/2$.

Using $\displaystyle Q=(-1/2,\pm\sqrt{3}/2)$ in (*), we get the desired tangent line equations.

Hello, atruenut!

There are two tangent lines from the point (0,1) to the circle $\displaystyle x^2 + (y+1)^2\:=\:1$
Find equations of these tangents.

| P * (0,1) /|\ . / | \ / | \ / | \ / | \ - - - - / - * * * - \ - - - - - - /* | *\ o | o A * * | * * * | * * * | * * * C *(0,-1) * * | * | * | * * | * * | * * * * |

The center of the circle is: $\displaystyle C(0,-1)$ Its radius is: $\displaystyle CA = 1$ A radius is perpendicular to a tangent at the point of tangency.

. . Hence: .$\displaystyle \angle CAP \:=\:90^o$

In right triangle $\displaystyle CAP$ we have: .$\displaystyle CA = 1,\;CP = 2$

. . Hence, $\displaystyle \Delta CAP$ is a 30-60 right triangle . . . $\displaystyle \angle APC = 30^o$ You should be able to determine the slope of $\displaystyle PA$ . . . right?

Reactions: bkarpuz

Q&a Wo

There are two tangent lines from the point (0 1)

zusammenhängende Posts

Werbung

NEUESTEN NACHRICHTEN

Toplisten

Werbung

Populer

Token Data

Werbung

Um

Legal

Hilfe

Sozial