If a tangent to the ellipse ${\mathrm{x}}^2 + 4{\mathrm{y}}^2 = 4$ meets the tangents at the extremities of its major axis at B and C, then the circle with BC as diameter passes through the point:

given ellipse $\frac{{\mathrm{x}}^2}{4}+\frac{{\mathrm{y}}^2}{1}=1$ 

equation of Tangent is 

$$\begin{gathered} \frac{\mathrm{x}}{2}\mathrm{cos\theta }+\frac{\mathrm{y}}{1}\mathrm{sin\theta }=1 \\ \mathrm{is} \mathrm{meets} the \tan gents \mathrm{x}=2 \& \mathrm{x}=-2 at \\ \mathrm{B}\left(2, \frac{1-\mathrm{cos\theta }}{\mathrm{sin\theta }}\right) \\ \mathrm{C}\left(-2, \frac{1+\mathrm{cos\theta }}{\mathrm{sin\theta }}\right) \end{gathered}$$

equation  of circle B, C as diameter is

 $$\begin{gathered} (\mathrm{x}-2) (\mathrm{x}+2)+\left(\mathrm{y}-\frac{1-\mathrm{cos\theta }}{\mathrm{sin\theta }}\right) \left(\mathrm{y}-\frac{1+\mathrm{cos\theta }}{\mathrm{sin\theta }}\right)=0 \\ \mathrm{put} \mathrm{y}=0 \\ {\mathrm{x}}^2-4+1=0 \\ {\mathrm{x}}^2=3 \Rightarrow \mathrm{x}=\pm \sqrt{3} \end{gathered}$$