1. Tangent to ellipse

Can some one help in finding the two lines that are tangent to the ellipse

2x2 – 4x + y2 + 1 = 0

Thank you...

2. 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:

$2x^2 -4x + y^2 + 1 = 0$ or
this
2x^2 -4x + y^2 + 1 = 0
would have been easier to understand

3. Tangent to ellipse

x2 and y2 should be read as to the power of 2. Further, the tangents lines pass through the origin.

My apologies...

4. Originally Posted by scarecrow13
Can some one help in finding the two lines that are tangent to the ellipse

2x2 – 4x + y2 + 1 = 0

Thank you...
Let $(x_0,y_0)$ be a point on the ellipse.

Thus, the derivative is $4x-4+2y\frac{\,dy}{\,dx}=0\implies \frac{\,dy}{\,dx}=\frac{2(1-x)}{y}$

Thus, $\left.\frac{\,dy}{\,dx}\right|_{(x_0,y_0)}=\frac{2 (1-x_0)}{y_0}$

Thus, the tangent line has the form $y-y_0=\left[\frac{2(1-x_0)}{y_0}\right](x-x_0)$

Since the lines pass through the origin, we see that ${\color{red}-y_0^2}=-2(x_0-x_0^2)$

What do we do now?

Recall that $2x^2-4x+y^2+1=0$. Thus, at the point $(x_0,y_0)$, we get $2x_0^2-4x_0+y_0^2+1=0\implies 2x_0^2-4x_0+1={\color{red}-y_0^2}$

Thus, substituting this into the mess we came up for the tangent lines, we see that $2x_0^2-4x_0+1=-2(x_0-x_0^2)\implies 1=2x_0\implies {\color{red}x_0=\tfrac{1}{2}}$

Substitution this into the expression for $-y_0^2$, we get that $2\left(\tfrac{1}{2}\right)^2-4\left(\tfrac{1}{2}\right)+1=-y_0^2\implies -\tfrac{1}{2}=-y_0^2\implies {\color{red}y_0=\pm\tfrac{\sqrt{2}}{2}}$

Thus, at the point $\left(\tfrac{1}{2},-\tfrac{\sqrt{2}}{2}\right)$, the tangent line has the equation $y+\tfrac{\sqrt{2}}{2}=-\sqrt{2}\left(x-\tfrac{1}{2}\right)\implies y=-\sqrt{2}x+\tfrac{\sqrt{2}}{2}-\tfrac{\sqrt{2}}{2}\implies\color{red}\boxed{y=-\sqrt{2}x}$

Thus, at the point $\left(\tfrac{1}{2},\tfrac{\sqrt{2}}{2}\right)$, the tangent line has the equation $y-\tfrac{\sqrt{2}}{2}=\sqrt{2}\left(x-\tfrac{1}{2}\right)\implies y=\sqrt{2}x-\tfrac{\sqrt{2}}{2}+\tfrac{\sqrt{2}}{2}\implies\co lor{red}\boxed{y=\sqrt{2}x}$

Does this make sense?

--Chris

5. tangent line to the ellipse

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