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