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Find the equation of the tangent and the normal to the following curves at the indicated points: 

\(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) at \((\sqrt{2}a, b)\)

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finding slope of the tangent by differentiating the curve

\(\frac{x}{a^2}-\frac{y}{b^2}\frac{dy}{dx}=0\)

\(\frac{dy}{dx}=\frac{xb^2}{ya^2}\)

m(tangent) at \((\sqrt{2}a,b)=\frac{\sqrt{2}ab^2}{2a^2}\)

normal is perpendicular to tangent so, m1m2 = – 1

m(normal) at \((\sqrt{2}a,b)=-\frac{ba^2}{\sqrt{2}ab^2}\)

equation of tangent is given by y – y1 = m(tangent)(x – x1)

\(y-b=\frac{\sqrt{2}ab^2}{ba^2}(x-\sqrt{2}a)\)

equation of normal is given by y – y1 = m(normal)(x – x1)

\(y-b=-\frac{ba^2}{\sqrt{2}ab^2}(x-\sqrt{2}a)\)

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