AP DSC School Assistant Math Model Test 2

Concept:The angle bisectors of a pair of lines given by $ax^{2} + 2hxy + by^{2} = 0$ satisfy $\frac{x^{2} - y^{2}}{a - b} = \frac{xy}{h}$.Explanation:The first equation is $x^{2} - 2pxy - y^{2} = 0$. Compare with $ax^{2} + 2hxy + by^{2} = 0$: here $a = 1$, $h = -p$, $b = -1$.Using the bisector formula:$$\frac{x^{2} - y^{2}}{1 - (-1)} = \frac{xy}{-p} \quad \Rightarrow \quad \frac{x^{2} - y^{2}}{2} = -\frac{xy}{p}$$Cross‑multiply: $p(x^{2} - y^{2}) = -2xy$Rearrange: $px^{2} + 2xy - py^{2} = 0$.This is the equation of the bisectors of the first pair. It must be identical to the second given equation $x^{2} - 2qxy - y^{2} = 0$ because the two pairs bisect each other’s angles.Comparing coefficients:$\frac{\text{coeff. of }x^{2}}{1} = \frac{\text{coeff. of }xy}{-2q} = \frac{\text{coeff. of }y^{2}}{-1}$From the bisector: coeff. of $x^{2} = p$, coeff. of $xy = 2$, coeff. of $y^{2} = -p$.Thus $\frac{p}{1} = \frac{2}{-2q} \quad \Rightarrow \quad p = -\frac{1}{q} \quad \Rightarrow \quad pq = -1 \quad \Rightarrow \quad pq + 1 = 0$.Answer:Option C: $pq + 1 = 0$