The equation of the normal to the curve \(x=a \cosh (t), y=b \sinh (t)\) at any point \(t\) is

The equation of the normal to the curve \(x=a \cosh (t), y=b \sinh (t)\) at any point \(t\) is
  1. \(a x+b y=a^2+b^2\)
  2. \(a x \operatorname{sech}(t)+b y \operatorname{cosech}(t)=a^2+b^2\)
  3. \(a x \operatorname{sech}(t)-b y \operatorname{cosech}(t)=a^2-b^2\)
  4. \(\frac{a x}{\sinh (t)}+\frac{b y}{\cosh (t)}=a^2+b^2\)

Solution

Given curve, \(x=a \cos \mathrm{h}(t)\) and \(y=b \sin \mathrm{h}(t)\) represents, \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) is a hyperbola and equation of normal at a point ' \(t\) ' is \(\begin{aligned} & \frac{x-a \cosh t}{\frac{a \cosh t}{a^2}}=-\frac{y-b \sinh t}{\frac{b \sin \mathrm{h} t}{b^2}} \\ \Rightarrow \quad & a x \operatorname{sech} t+b y \operatorname{cosech} t=a^2+b^2 \end{aligned}\)

Asked in: AP EAMCET 2020 (18 Sep Shift 1)

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