Q.An amount of Rs.,65{,}000 is invested in three bonds at the rates of 6%, 8% and 9% per annum respectively. The total annual income is Rs.,4{,}800. The income from the third bond is Rs.,600 more than that from the second bond. Determine the amount invested in each bond. (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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