Find the equation of the tangent and the normal to the following curve

1.	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 (a sec θ, b tan θ)
Answer» finding the 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 (a secθ , b tanθ ) = \(\frac{b}{a\sin\theta}\) normal is perpendicular to tangent so, m₁m₂ = – 1 m(normal) at (a secθ , b tanθ ) = \(-\frac{a\sin\theta}{b}\) equation of tangent is given by y – y₁ = m(tangent)(x – x₁) \(y-b\tan\theta=\frac{b}{a\sin\theta}(x-a\sec\theta)\) equation of normal is given by y – y₁ = m(normal)(x – x₁) \(y-b\tan\theta=-\frac{a\sin\theta}{b}(x-a\sec\theta)\)

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 (a sec θ, b tan θ)

Answer»

finding the 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 (a secθ , b tanθ ) = \(\frac{b}{a\sin\theta}\)

normal is perpendicular to tangent so, m₁m₂ = – 1

m(normal) at (a secθ , b tanθ ) = \(-\frac{a\sin\theta}{b}\)

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

\(y-b\tan\theta=\frac{b}{a\sin\theta}(x-a\sec\theta)\)

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

\(y-b\tan\theta=-\frac{a\sin\theta}{b}(x-a\sec\theta)\)