If a tangent to the ellipse x^2+4y^2=4 meets the tangents at the extremities of its major …

Question

If a tangent to the ellipse $x^2+4y^2=4$ 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

Accepted Answer

$\dfrac{x^2}4+y^2=1$. Tangent at $(2\cos\theta,\sin\theta)$: $\dfrac{x\cos\theta}2+y\sin\theta=1$. At $x=2$: $y_B=\dfrac{1-\cos\theta}{\sin\theta}$; at $x=-2$: $y_C=\dfrac{1+\cos\theta}{\sin\theta}$.

$$y_By_C=\frac{1-\cos^2\theta}{\sin^2\theta}=1.$$

For a point $(x_0,0)$ on the circle with diameter $B(2,y_B),C(-2,y_C)$: $(2-x_0)(-2-x_0)+y_By_C=0\Rightarrow x_0^2-4+1=0\Rightarrow x_0=\pm\sqrt3$.

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