Question:

To solve this problem, let's start by considering the given pair of straight lines $ax^2 + 2hy + by^2 = 0$. We can rewrite this equation as $$\begin{aligned} ax^2 + 2hy + by^2 &= 0 \\ by^2 + 2hy + ax^2 &= 0 \end{aligned}$$ This equation represents a pair of straight lines. Since the line through point A on the x-axis is parallel to the y-axis, its equation is of the form $x = k$, where $k$ is a constant. Step 1: Find the points of intersection between the line $x = k$ and the pair of straight lines $ax^2 + 2hy + by^2 = 0$. Substituting $x = k$ into the equation of the pair of straight lines, we get $$\begin{aligned} ak^2 + 2hy + by^2 &= 0 \\ by^2 + 2hy + ak^2 &= 0 \end{aligned}$$ This is a quadratic equation in $y$, which can be factored as $$\begin{aligned} (b(y - y_1)(y - y_2)) &= 0 \\ y^2 + \frac{2h}{b}y + \frac{ak^2}{b} &= 0 \end{aligned}$$ Using the quadratic formula, we can find the roots of this equation as $$\begin{aligned} y_{1,2} &= \frac{-\frac{2h}{b} \pm \sqrt{\left(\frac{2h}{b}\right)^2 - 4\cdot1\cdot\frac{ak^2}{b}}}{2} \\ y_{1,2} &= \frac{-\frac{2h}{b} \pm \sqrt{\frac{4h^2}{b^2} - \frac{4ak^2}{b}}}{2} \\ y_{1,2} &= \frac{-\frac{2h}{b} \pm \frac{\sqrt{4h^2 - 4abk^2}}{b}}{2} \\ y_{1,2} &= \frac{-h \pm \sqrt{h^2 - abk^2}}{b} \end{aligned}$$ Step 2: Since $AB = BC$, the midpoint of the line segment $BC$ must coincide with point $A$. The x-coordinate of $A$ is $k$, and the y-coordinate is $0$. The midpoint of $BC$ can be found as $$\begin{aligned} y_{mid} &= \frac{y_1 + y_2}{2} \\ y_{mid} &= \frac{\frac{-h + \sqrt{h^2 - abk^2}}{b} + \frac{-h - \sqrt{h^2 - abk^2}}{b}}{2} \\ y_{mid} &= \frac{-2h}{2b} \\ y_{mid} &= \frac{-h}{b} \end{aligned}$$ Since $y_{mid} = 0$, we have $$\begin{aligned} \frac{-h}{b} &= 0 \\ -h &= 0 \end{aligned}$$ This implies $h = 0$, but this is not possible since $h$ is a non-zero constant. Therefore, we must reconsider our approach. Step 3: Reconsidering the equation $ax^2 + 2hy + by^2 = 0$, we can rewrite it as $$\begin{aligned} by^2 + 2hy + ax^2 &= 0 \\ y^2 + \frac{2h}{b}y + \frac{a}{b}x^2 &= 0 \end{aligned}$$ For the line $x = k$, we substitute $x = k$ into the equation above, giving $$\begin{aligned} y^2 + \frac{2h}{b}y + \frac{ak^2}{b} &= 0 \end{aligned}$$ Since $AB = BC$, the roots of this quadratic equation must be equal in magnitude but opposite in sign. This implies that the sum of the roots is zero, i.e., $$\begin{aligned} y_1 + y_2 &= 0 \\ \frac{-\frac{2h}{b}}{1} &= 0 \\ -2h &= 0 \end{aligned}$$ However, this would imply $h = 0$, which is not possible. Therefore, we reconsider the condition for $AB = BC$. Step 4: The condition $AB = BC$ implies that the line $x = k$ is the perpendicular bisector of the line segment $BC$. Since the line $x = k$ is parallel to the y-axis, the x-coordinate of the midpoint of $BC$ must be $k$. The equation of the pair of straight lines can be rewritten as $$\begin{aligned} ax^2 + 2hy + by^2 &= 0 \\ ax^2 + 2hy + by^2 - ab &= 0 + ab \\ ax^2 - ab + 2hy + by^2 &= ab \\ a(x^2 - b) + b(y^2 + 2\frac{h}{b}y) &= ab \\ a(x^2 - b) + b(y^2 + 2\frac{h}{b}y + \frac{h^2}{b^2}) &= ab + \frac{h^2}{b} \\ a(x^2 - b) + b(y + \frac{h}{b})^2 &= ab + \frac{h^2}{b} \end{aligned}$$ For the line $x = k$, we substitute $x = k$ into the equation above, giving $$\begin{aligned} a(k^2 - b) + b(y + \frac{h}{b})^2 &= ab + \frac{h^2}{b} \\ b(y + \frac{h}{b})^2 &= ab + \frac{h^2}{b} - a(k^2 - b) \\ b(y + \frac{h}{b})^2 &= ab + \frac{h^2}{b} - ak^2 + ab \\ b(y + \frac{h}{b})^2 &= \frac{h^2}{b} + 2ab - ak^2 \end{aligned}$$ Since $AB = BC$, the roots of this quadratic equation must be equal in magnitude but opposite in sign. This implies that the sum of the roots is zero, i.e., $$\begin{aligned} y_1 + y_2 &= -2\cdot\frac{h}{b} \\ -2\cdot\frac{h}{b} &= 0 \end{aligned}$$ However, this would imply $h = 0$, which is not possible. Therefore, we reconsider the condition for $AB = BC$. Step 5: The condition $AB = BC$ implies that the line $x = k$ is the perpendicular bisector of the line segment $BC$. Since the line $x = k$ is parallel to the y-axis, the x-coordinate of the midpoint of $BC$ must be $k$. For the equation $ax^2 + 2hy + by^2 = 0$, the roots can be found as $$\begin{aligned} y_{1,2} &= \frac{-2h \pm \sqrt{4h^2 - 4abx^2}}{2b} \\ y_{1,2} &= \frac{-2h \pm 2\sqrt{h^2 - abx^2}}{2b} \\ y_{1,2} &= \frac{-h \pm \sqrt{h^2 - abx^2}}{b} \end{aligned}$$ The midpoint of $BC$ can be found as $$\begin{aligned} y_{mid} &= \frac{y_1 + y_2}{2} \\ y_{mid} &= \frac{\frac{-h + \sqrt{h^2 - abx^2}}{b} + \frac{-h - \sqrt{h^2 - abx^2}}{b}}{2} \\ y_{mid} &= \frac{-2h}{2b} \\ y_{mid} &= \frac{-h}{b} \end{aligned}$$ Since the line $x = k$ is parallel to the y-axis, $y_{mid} = 0$. However, this would imply $h = 0$, which is not possible. Therefore, we reconsider the equation $ax^2 + 2hy + by^2 = 0$. Step 6: The equation $ax^2 + 2hy + by^2 = 0$ represents a pair of straight lines. For the line $x = k$ to be the perpendicular bisector of $BC$, we must have $$\begin{aligned} h^2 - abk^2 &= \frac{h^2}{4} \\ 4h^2 - 4abk^2 &= h^2 \\ 3h^2 &= 4abk^2 \\ 3h^2 &= 4ab\cdot k^2 \end{aligned}$$ However, we are given that $AB <div class="key-concept"><strong>Key Concept:</strong> using the properties of the perpendicular bisector of a line segment and the relationship between the roots of a quadratic equation</div> <div class="trap-box"><strong>Trap:</strong> incorrectly assuming that the sum of the roots of the quadratic equation is zero, leading to the conclusion that h = 0, which is not possible</div>