If the points of intersection of the ellipse | vedclass

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(A) Given the curve $y^{2}=3x^{2}$ and the circle $x^{2}+y^{2}=4b$.Substituting $y^{2}=3x^{2}$ into the circle equation: $x^{2}+3x^{2}=4b \implies 4x^

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(A) Given the curve $y^{2}=3x^{2}$ and the circle $x^{2}+y^{2}=4b$.Substituting $y^{2}=3x^{2}$ into the circle equation: $x^{2}+3x^{2}=4b \implies 4x^{2}=4b \implies x^{2}=b$.Then $y^{2}=3b$.Now,substitute $x^{2}=b$ and $y^{2}=3b$ into the ellipse equation $\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1$:$\frac{b}{16}+\frac{3b}{b^{2}}=1$$\frac{b}{16}+\frac{3}{b}=1$Multiply by $16b$: $b^{2}+48=16b$$b^{2}-16b+48=0$$(b-12)(b-4)=0$Since $b > 4$,we have $b=12$.

If the points of intersection of the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1$ and the circle $x^{2}+y^{2}=4b$ (where $b > 4$) lie on the curve $y^{2}=3x^{2}$,then $b$ is equal to:

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