The questions reads
Find equations for two lines through the origin that are tangent to the curve $\displaystyle x^2-4x+y^2+3=0$ ,, Not sure what the question is asking..
Any help?
$\displaystyle (x- 2)^2= x^2- 4x+ 4$ so that $\displaystyle x^2- 4x+ y^2+ 3= x^2- 4x+ 4- 4+ y^2+ 3= (x-2)^2+ y^2- 1= 0$ so that $\displaystyle (x- 2)^2+ y^2= 1$. That is the equation of a circle with center (2, 0) and radius 1. The problem ask for you to find two equations through the origin (so y= mx for some m) that are tangent to the circle. It should be clear, geometrically, that one is above the x-axis and the other below.
One way to do this is to argue that the line must be tangent to the circle and so replacing y by mx, [must](x- 2)^2+ (mx)^2= 1[/tex], must have a double root. That is a quadratic equation so must have zero, one, or two solutions. Since a tangent touches the circle, it cannot have 0 solutions. Since a tangent does not pass through the circle, it does not cross it twice and so that quadratic equation must have only one solution. You can set the discriminant of the equation equal to 0 to determine m.
Another way to do this is to use the fact that a tangent to a circle is perpendicular to a radius. So the origin, (0, 0), the center of the circle, (2, 0), and the point, (x, y), where the line is tangent to the circle form a right triangle with the x-axis, from (0, 0) to (3, 0) as hypontenuse. That is, $\displaystyle x^2+y^2= 4+ 1= 5$ since the distance from (0, 0) to (2, 0) is 2 and the circle has radius 1. That, together with the equation of the circle gives you two equations to solve for x and y. Once you know the point, (x, y), where the line is tangent to the circle, as well as (0, 0) you can write down the equation of the line.
If $\displaystyle (a,b)$ is on that curve we know that $\displaystyle a^2-4a+b^2+3=0$.
We know that the slope at each point is $\displaystyle y'=\frac{2-x}{y}$.
If $\displaystyle a\ne 0$ then any line through $\displaystyle (0,0)~\&~(a,b)$ has slope $\displaystyle \frac{b}{a}$.
Now, you want to find point $\displaystyle (a,b)$ on the curve that has slope $\displaystyle y'=\frac{2-a}{b}=\frac{b}{a}$
Ameer
Problems like the one you posted yesterday can be easily solved if you consider the Polar line of the conic.
the Polar line is the line that connects the two points of tangency. it has the same form as the equation of the tangent to a circle .
In your case the conic is a circle of the form x^2+y^2+Ax +By +C = 0 where A=-4, B= 0 and C = 3.
The point you are reffered to is the origin O(x1,y1)=(0,0).
In this case the equation of the polar line to the conic has the form : xx1+yy1+A(x+x1)/2+B(y+y1)/2+C=0 and by substitution you get x = 3/2 since x1=y1=0.
now solve the system of the conic and the polar line to find the coordinates of the points of tangency . the rest is grade 8 math....
I attach a figure for easy reference.
MINOAS
now I got this equation $\displaystyle y^2=2x-x^2$ or $\displaystyle b^2=2a-a^2$ if I follow Plato.
Isn't that is the answer?
But I see in my book they went a step ahead, substituted $\displaystyle b^2$ with $\displaystyle 2a-a^2$ and came with the result
$\displaystyle b=(\sqrt3/3)a$ and $\displaystyle b=-(\sqrt3/3)a$
Is that even necessary? I thought $\displaystyle b^2=2a-a^2$ is the equation and should be the answer isn't it?
Or at bets I could remove the squares $\displaystyle b=\sqrt{2a-a^2}$