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Question

(A) Substitute $y^2=3x^2$ into both conic equations.For the first conic: $x^2+3x^2=4b$ $\Rightarrow 4x^2=4b$ $\Rightarrow x^2=b$.For the second conic:

Vedclass · Accepted Answer

$432$

Vedclass · Answer

(A) Substitute $y^2=3x^2$ into both conic equations.For the first conic: $x^2+3x^2=4b$ $\Rightarrow 4x^2=4b$ $\Rightarrow x^2=b$.For the second conic: $\frac{x^2}{16}+\frac{3x^2}{b^2}=1 \Rightarrow \frac{b}{16}+\frac{3}{b}=1$.Multiplying by $16b$,we get $b^2+48=16b \Rightarrow b^2-16b+48=0$.Factoring gives $(b-12)(b-4)=0$,so $b=12$ or $b=4$.If $b=4$,the conics are $x^2+y^2=16$ and $\frac{x^2}{16}+\frac{y^2}{16}=1$,which coincide. Thus,$b=12$.For $b=12$,$x^2=12 \Rightarrow x = \pm 2\sqrt{3}$ and $y^2=3(12)=36 \Rightarrow y = \pm 6$.The intersection points are $(\pm 2\sqrt{3}, \pm 6)$.The rectangle has vertices $(\pm 2\sqrt{3}, \pm 6)$,so its width is $4\sqrt{3}$ and height is $12$.Area $= (4\sqrt{3}) 	imes 12 = 48\sqrt{3}$.The value requested is $3\sqrt{3} 	imes (48\sqrt{3}) = 3 	imes 48 	imes 3 = 432$.

If the points of intersection of two distinct conics $x^2+y^2=4b$ and $\frac{x^2}{16}+\frac{y^2}{b^2}=1$ lie on the curve $y^2=3x^2$,then $3\sqrt{3}$ times the area of the rectangle formed by the intersection points is............................

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