Common tangent to a circle and ellipse Hey guys i am noy able to solve this problem.So please do help me in solving this.The equation of common tangent to ellipse 
\begin{equation*}
x^2 +2y^2=1
\end{equation*}
and circle 
\begin{equation*}
x^2 +y^2=\frac{2}{3}
\end{equation*}
is?
 A: Any tangent on the circle at $(\sqrt{2/3}\cos t,\sqrt{2/3}\sin t)$
$x(\sqrt{2/3}\cos t)+y(\sqrt{2/3}\sin t)=2/3$
$\iff x(\cos t)+y(\sin t)=\sqrt{2/3}\  \  \ \  (1)$
Any tangent on the ellipse at $(\cos u,\sqrt{1/2}\sin u)$
$x(\cos u)/2+y(\sqrt{1/2}\sin u)=1/2 \  \  \ \  (2)$
We need $(1),(2)$ to be the same straight line 
$$\implies\dfrac{\cos t}{\cos u}=\dfrac{\sin t}{\sqrt{1/2}\sin u}=\dfrac{\sqrt{2/3}}{1/2}$$
Find $\cos t,\sin t$ and eliminate them using $\cos^2+\sin^2t=1$ to find $\cos u,\sin u$
A: Hint: Use the auxiliary circle property of ellipses: A perpendicular to a tangent from a focal point will meet the tangent on the auxiliary circle.
The auxiliary circle of the ellipse $x^2 + 2y^2 = 1$ is the circle $x^2 + y^2 = 1$. You get three line segments perpendicular to the tangent - one from each focus of the ellipse and one from the center. Using the known lengths and the law of sines, you can work out the slope and intercept of the tangent line.
A: The equation of tangent on the ellipse is given by:
$$x x_1 + 2y y_1 = 1$$
The perpendicular distance of the line from the centre of circle (0, 0)
$= \frac 1{\sqrt{x_1^2 + 4y_1^2}}$
This distance is equal to the radius of the circle as the line should also be tangent to the circle:
$$\frac 1{\sqrt{x_1^2 + 4y_1^2}} = \sqrt \frac23$$
Since $(x_1, y_1)$ lie on the ellipse, $x_1^2 + 2y_1^2 = 1$; These give $y_1 = \pm \frac12$ and $x_1 = \pm \sqrt \frac12$.
Therefore, the tangents are: $$ \frac{\pm x}{\sqrt 2} \pm y = 1$$

Way 2: Let the common tangent be $y = mx + c$.
It meets the ellipse at only one point:
$$mx + c = \sqrt{\frac{1-x^2}2}$$
$$(2m^2 + 1)x^2 + 4mcx + 2c^2 - 1 = 0$$
This quadratic has equal roots so discriminant = 0
$$2m^2 + 1 = 2c^2$$
Similarly for the circle,
$$mx + c = \sqrt{\frac23 - x^2}$$
Simplifying and setting discriminant to 0 gives:
$$2m^2 + 2 = 3c^2$$
Solving for m and c gives
$$ \frac{\pm x}{\sqrt 2} \pm y = 1$$
Desmos graph
A: A method that would also work for other pairs of conics: $$x^2 +2y^2-1=0 , x^2 +y^2-\frac23=0$$ have dual conics $$X^2+\frac12Y^2-1=0, \frac23X^2+\frac23Y^2-1=0$$ which intersect in $$(X,Y)=(-\frac1{\sqrt2},-1),(-\frac1{\sqrt2},1),(\frac1{\sqrt2},-1),(\frac1{\sqrt2},1)$$

and by substituting into the equation of the universal line $xX+yY+1=0$ we get the four common tangent lines $$-\frac1{\sqrt2}x-y+1=0,-\frac1{\sqrt2}x+y+1=0,\frac1{\sqrt2}x,-y+1=0,\frac1{\sqrt2}x+y+1=0.$$

