Why there are not 4 tangents from a point to a hyperbola or ellipse?

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The equation of a tangent to a ellipse (x²/a²)+(y²/b²)=1 can be written as y=mx±√(a²m²+b²) ,where m is the slope of the tangent, now if this tangent passes through a specific point outside the ellipse, we can get a quadratic equation in terms of m, solving which will give us 2 values of m for the 2 tangents drawn to the ellipse from that point, but, there is also a "±" term before √(a²m²+b²) so each value of m would give 2 tangents corresponding to it? Then why we say that only 2 tangents can be drawn from a point to a ellipse?

There is a similarly titled question that already exists, but, does not specifically clarify the point I am asking. Thank you.

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Standard Coordinate Geometry approach:

To find the equations of the tangents from an external point $(\alpha, \beta)$ to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. enter image description here We let the equation of the tangent be $y=mx+c$.

Putting $y=mx+c$ into $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and using $\Delta = 0$, we get $$c^2=a^2m^2+b^2.$$

To find the values of $m$ and $c$, we solve

$$c^2=a^2m^2+b^2 \tag{1}$$ and $$\beta=m\alpha+c \tag{2}$$

From $(1), (2)$, we have $$(\beta - m \alpha)^2=a^2m^2+b^2.$$

This is a quadratic equation in $m$ and hence in general has $2$ solutions in $m$.

But for each value of $m$ obtained, equation $(2)$, i.e. $$c=\beta - m \alpha$$ will give only one single value of $c$.