Can some one help in finding the two lines that are tangent to the ellipse
2x2 – 4x + y2 + 1 = 0
Thank you...
There are infinitely many lines that are tangent to an ellipse, so which 2 do you want?
Also it would be much easier to understand your problem and equation if you learn latex or at least use the proper notions of symbols like ^ means to the power of...
For instance this:
$\displaystyle 2x^2 -4x + y^2 + 1 = 0 $ or
this
2x^2 -4x + y^2 + 1 = 0
would have been easier to understand
Let $\displaystyle (x_0,y_0)$ be a point on the ellipse.
Thus, the derivative is $\displaystyle 4x-4+2y\frac{\,dy}{\,dx}=0\implies \frac{\,dy}{\,dx}=\frac{2(1-x)}{y}$
Thus, $\displaystyle \left.\frac{\,dy}{\,dx}\right|_{(x_0,y_0)}=\frac{2 (1-x_0)}{y_0}$
Thus, the tangent line has the form $\displaystyle y-y_0=\left[\frac{2(1-x_0)}{y_0}\right](x-x_0)$
Since the lines pass through the origin, we see that $\displaystyle {\color{red}-y_0^2}=-2(x_0-x_0^2)$
What do we do now?
Recall that $\displaystyle 2x^2-4x+y^2+1=0$. Thus, at the point $\displaystyle (x_0,y_0)$, we get $\displaystyle 2x_0^2-4x_0+y_0^2+1=0\implies 2x_0^2-4x_0+1={\color{red}-y_0^2}$
Thus, substituting this into the mess we came up for the tangent lines, we see that $\displaystyle 2x_0^2-4x_0+1=-2(x_0-x_0^2)\implies 1=2x_0\implies {\color{red}x_0=\tfrac{1}{2}}$
Substitution this into the expression for $\displaystyle -y_0^2$, we get that $\displaystyle 2\left(\tfrac{1}{2}\right)^2-4\left(\tfrac{1}{2}\right)+1=-y_0^2\implies -\tfrac{1}{2}=-y_0^2\implies {\color{red}y_0=\pm\tfrac{\sqrt{2}}{2}}$
Thus, at the point $\displaystyle \left(\tfrac{1}{2},-\tfrac{\sqrt{2}}{2}\right)$, the tangent line has the equation $\displaystyle y+\tfrac{\sqrt{2}}{2}=-\sqrt{2}\left(x-\tfrac{1}{2}\right)\implies y=-\sqrt{2}x+\tfrac{\sqrt{2}}{2}-\tfrac{\sqrt{2}}{2}\implies\color{red}\boxed{y=-\sqrt{2}x}$
Thus, at the point $\displaystyle \left(\tfrac{1}{2},\tfrac{\sqrt{2}}{2}\right)$, the tangent line has the equation $\displaystyle y-\tfrac{\sqrt{2}}{2}=\sqrt{2}\left(x-\tfrac{1}{2}\right)\implies y=\sqrt{2}x-\tfrac{\sqrt{2}}{2}+\tfrac{\sqrt{2}}{2}\implies\co lor{red}\boxed{y=\sqrt{2}x}$
Does this make sense?
--Chris
I need help with this one.
Find the tangent line(s) to the ellipse x^2 + 2y^2 +4x = 5 at the point (1,0)
I differentiated the equation implicitly to get this: (-2x-4)/4y
And thats where i got stuck. cuz i get a ZERO in the denoinator.
well,, using analytic math I got this equation (not sure if its correct either): x+5/3=0
but i HAVE TO get there using differentiation
the teacher said something like this is gonna appear on the test, so please help