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(A) The given equation is $x^2+4xy+y^2=1$. This is a conic section of the form $Ax^2+2Hxy+By^2=C$. Here,$A=1, H=2, B=1, C=1$. The invariants under rot

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(A) The given equation is $x^2+4xy+y^2=1$. This is a conic section of the form $Ax^2+2Hxy+By^2=C$. Here,$A=1, H=2, B=1, C=1$. The invariants under rotation of axes are $A+B$ and $AB-H^2$. Sum of coefficients: $A+B = 1+1 = 2$. Discriminant: $AB-H^2 = (1)(1) - (2)^2 = 1-4 = -3$. After rotation,the equation becomes $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$,which can be written as $\frac{1}{a^2}x^2 - \frac{1}{b^2}y^2 = 1$. Comparing this with the standard form $A'x^2+B'y^2=C'$,we have $A' = \frac{1}{a^2}$,$B' = -\frac{1}{b^2}$,and $C'=1$. The invariants are: $A'+B' = \frac{1}{a^2} - \frac{1}{b^2} = 2$ $A'B' = (\frac{1}{a^2})(-\frac{1}{b^2}) = -3$ From these,we have $\frac{1}{a^2} - \frac{1}{b^2} = 2$ and $\frac{1}{a^2b^2} = 3$. We need to find $\sqrt{\frac{a^2+b^2}{a^2}} = \sqrt{1+\frac{b^2}{a^2}}$. Since $\frac{1}{a^2b^2} = 3$,we have $b^2 = \frac{1}{3a^2}$. Substituting this into the first invariant: $\frac{1}{a^2} - 3a^2 = 2$. Let $u = \frac{1}{a^2}$. Then $u - 3/u = 2 \implies u^2 - 2u - 3 = 0$. $(u-3)(u+1) = 0$. Since $u = 1/a^2 > 0$,we have $u = 3$,so $a^2 = 1/3$. Then $b^2 = 1/(3 	imes 1/3) = 1$. Thus,$\sqrt{1+\frac{1}{1/3}} = \sqrt{1+3} = \sqrt{4} = 2$.

By rotating the axes about the origin in an anti-clockwise direction by a certain angle,if the equation $x^2+4xy+y^2=1$ is transformed to $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$,then find the value of $\sqrt{\frac{a^2+b^2}{a^2}}$.

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