AP DSC School Assistant Math Model Test 2

Concept:The general equation of a cone with vertex at the origin is $ax^{2}+by^{2}+cz^{2}+2fyz+2gzx+2hxy=0$. When the cone passes through the three coordinate axes, the points $(x,0,0)$, $(0,y,0)$, $(0,0,z)$ satisfy the equation, forcing $a=b=c=0$. Thus the cone reduces to $fyz+gzx+hxy=0$. Any line through the vertex that lies entirely on the cone (a generator) must have its direction ratios satisfy this equation.Explanation:Let the cone be $ax^{2}+by^{2}+cz^{2}+2fyz+2gzx+2hxy=0$. Since the coordinate axes are on the cone, substitute $(x,0,0)$: $ax^{2}=0$ ⇒ $a=0$. Similarly, $(0,y,0)$ gives $b=0$, and $(0,0,z)$ gives $c=0$. Hence the equation becomes $2fyz+2gzx+2hxy=0$, or $fyz+gzx+hxy=0$ — (1).The two given lines are generators. Their direction ratios are $(1,-2,3)$ and $(3,-1,1)$. Substituting into (1) for the first line: $f(-2)(3)+g(3)(1)+h(1)(-2)=0$ ⇒ $-6f+3g-2h=0$ — (2).For the second line: $f(-1)(1)+g(1)(3)+h(3)(-1)=0$ ⇒ $-f+3g-3h=0$ — (3).Solve (2) and (3) by ratios method. Write the system as:$-6f+3g-2h=0$$-f+3g-3h=0$Treat as homogeneous linear equations in $f,g,h$. Cross-multiply the coefficients:$\frac{f}{\begin{vmatrix}3 & -2\\3 & -3\end{vmatrix}} = \frac{-g}{\begin{vmatrix}-6 & -2\\-1 & -3\end{vmatrix}} = \frac{h}{\begin{vmatrix}-6 & 3\\-1 & 3\end{vmatrix}}$Compute determinants:For $f$: $(3)(-3)-(-2)(3) = -9+6=-3$For $g$ (with negative sign): $-\left[(-6)(-3)-(-2)(-1)\right] = -\left[18-2\right]=-16$For $h$: $(-6)(3)-(3)(-1)=-18+3=-15$Thus $\frac{f}{-3}=\frac{g}{-16}=\frac{h}{-15}=k$ (say). So $f=-3k$, $g=-16k$, $h=-15k$.Plug into (1): $(-3k)yz+(-16k)zx+(-15k)xy=0$. Divide by $-k$ (nonzero): $3yz+16zx+15xy=0$.Answer:Option C: $3yz+16zx+15xy=0$.