Q.A boy is walking along the path $y=ax^2+bx+c$ through the points $(-6,8), (-2,-12)$, and $(3,8)$. He wants to meet his friend at $P(7,60)$. Will he meet his friend? (Use Gaussian elimination method.)

(a) Sets up and Gaussian-eliminates a 3-equation linear system for $a,b,c$ in $y=ax^2+bx+c$, then checks whether $(7,60)$ lies on the resulting parabola; (b) differentiates both conics implicitly and shows the product of their slopes is $-1$ at every intersection point. Both alternatives answered below.

(a) Path through 3 points — Gaussian elimination

1. Set up equations. Substituting each point into $y=ax^2+bx+c$:

• $(-6,8)$: $36a-6b+c=8$
• $(-2,-12)$: $4a-2b+c=-12$
• $(3,8)$: $9a+3b+c=8$

2. Eliminate $c$. (Eq.1) $-$ (Eq.2): $32a-4b=20\ \Rightarrow\ 8a-b=5$ … (i). (Eq.3) $-$ (Eq.2): $5a+5b=20\ \Rightarrow\ a+b=4$ … (ii).

3. Solve the reduced $2\times2$ system. From (i): $b=8a-5$. Substitute into (ii): $a+(8a-5)=4\Rightarrow9a=9\Rightarrow a=1$. Then $b=8(1)-5=3$.

4. Back-substitute for $c$. Using Eq.2: $4(1)-2(3)+c=-12\Rightarrow4-6+c=-12\Rightarrow c=-10$.

5. Path equation. $y=x^2+3x-10$. Check against all three original points: $(-6,8)$: $36-18-10=8$✓; $(-2,-12)$: $4-6-10=-12$✓; $(3,8)$: $9+9-10=8$✓.

6. Test the friend's point $(7,60)$. $y(7)=7^2+3(7)-10=49+21-10=60$ — exactly matches. Yes, the boy will meet his friend at $P(7,60)$, since $(7,60)$ lies on his path.

(b) Orthogonal intersection of $x^2+4y^2=8$ and $x^2-2y^2=4$

1. Find the points of intersection. From the hyperbola, $x^2=4+2y^2$. Substitute into the ellipse: $(4+2y^2)+4y^2=8\Rightarrow6y^2=4\Rightarrow y^2=\dfrac23$. Then $x^2=4+2\left(\dfrac23\right)=4+\dfrac43=\dfrac{16}{3}$.

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