How the Opening Direction of a Family of Parabolas Changes
When studying a family of parabolas, one of the most common questions is how the parabola changes as the parameter \(\lambda\) varies. In particular, we may want to know when the parabola opens upward, when it opens downward, and which values of the parameter cause a change in its orientation.
To answer these questions, consider the general equation of a family of parabolas:
$$ Ax^2 + Bx + Cy + D = 0 $$
where the coefficients depend on the parameter \(\lambda\):
$$ A = a_1 + \lambda a_2 $$
$$ B = b_1 + \lambda b_2 $$
$$ C = 1 + \lambda $$
$$ D = c_1 + \lambda c_2 $$
To determine the opening direction, rewrite the equation in explicit form by solving for \(y\):
$$ y = -\frac{A}{C}x^2 - \frac{B}{C}x - \frac{D}{C} $$
provided that \(C \neq 0\).
At this point, the problem becomes very simple. The opening direction depends entirely on the sign of the coefficient of \(x^2\):
$$ -\frac{A}{C} $$
- If \(-\frac{A}{C} > 0\), the parabola opens upward.
- If \(-\frac{A}{C} < 0\), the parabola opens downward.
This is the most direct and reliable method because it automatically takes into account the effect of both coefficients \(A\) and \(C\).
If \(A=0\), the equation is no longer quadratic in \(x\), so it does not represent an ordinary parabola. Instead, the curve becomes a degenerate member of the family.
Note: Some textbooks analyze the signs of \(A\) and \(C\) separately. While that approach works, checking the sign of \(-\frac{A}{C}\) is usually faster and provides the same result in a single step.
An Example
Consider the family of parabolas:
$$ (y - x^2 + 3x - 2) + \lambda (y + x^2 - 5x + 4) = 0 $$
Begin by expanding the expression:
$$ y - x^2 + 3x - 2 + \lambda y + \lambda x^2 - 5\lambda x + 4\lambda = 0 $$
Combining like terms gives:
$$ (1+\lambda)y + (\lambda-1)x^2 + (3-5\lambda)x + (-2+4\lambda) = 0 $$
This equation can be written in the form:
$$ Ax^2 + Bx + Cy + D = 0 $$
with:
$$ A = \lambda - 1 $$
$$ B = 3 - 5\lambda $$
$$ C = 1 + \lambda $$
$$ D = -2 + 4\lambda $$
Solving for \(y\) yields:
$$ y = -\frac{\lambda - 1}{1+\lambda}x^2 - \frac{3-5\lambda}{1+\lambda}x - \frac{-2+4\lambda}{1+\lambda} $$
Therefore, the opening direction depends on the sign of:
$$ -\frac{\lambda - 1}{1+\lambda} $$
The critical values occur when either the numerator or the denominator is zero:
$$ \lambda - 1 = 0 \qquad \Rightarrow \qquad \lambda = 1 $$
$$ 1 + \lambda = 0 \qquad \Rightarrow \qquad \lambda = -1 $$
These values divide the real line into three intervals:
- \((-\infty,-1)\)
- \((-1,1)\)
- \((1,\infty)\)
Now evaluate the sign of \(-\frac{\lambda - 1}{1+\lambda}\) in each interval:
- For \(\lambda < -1\), the expression is negative, so the parabola opens downward.
- For \(-1 < \lambda < 1\), the expression is positive, so the parabola opens upward.
- For \(\lambda > 1\), the expression is negative again, so the parabola opens downward.
The result can be summarized as follows:
- For \(\lambda < -1\), the parabola opens downward.
- For \(-1 < \lambda < 1\), the parabola opens upward.
- For \(\lambda > 1\), the parabola opens downward.
What Happens at the Critical Values?
The values \(\lambda=-1\) and \(\lambda=1\) are special because the equation no longer represents an ordinary parabola.
When \(\lambda=1\), the coefficient of \(x^2\) becomes zero. The equation simplifies to:
$$ 2y - 2x + 2 = 0 $$
or:
$$ y = x - 1 $$
In this case, the parabola degenerates into a straight line passing through the two base points of the family.
When \(\lambda=-1\), the coefficient of \(y\) becomes zero, and the equation reduces to:
$$ -2x^2 + 8x - 6 = 0 $$
Dividing by \(-2\):
$$ x^2 - 4x + 3 = 0 $$
Factoring the quadratic gives:
$$ (x-1)(x-3)=0 $$
Therefore, the degenerate conic consists of the two vertical lines:
$$ x=1 $$
$$ x=3 $$
These lines are parallel to the axis of symmetry of the parabolas in the family.
Key Point: For a family of parabolas written in the form \(Ax^2+Bx+Cy+D=0\), the easiest way to determine whether the parabola opens upward or downward is to study the sign of \(-\frac{A}{C}\). Once the critical values have been identified, a simple sign analysis reveals how the opening direction changes as the parameter varies.
The same method can be used for any one-parameter family of parabolas.
