AP EAMCET · Maths · Hyperbola
The equation of the normal to the curve \(x=a \cosh (t), y=b \sinh (t)\) at any point \(t\) is
- A \(a x+b y=a^2+b^2\)
- B \(a x \operatorname{sech}(t)+b y \operatorname{cosech}(t)=a^2+b^2\)
- C \(a x \operatorname{sech}(t)-b y \operatorname{cosech}(t)=a^2-b^2\)
- D \(\frac{a x}{\sinh (t)}+\frac{b y}{\cosh (t)}=a^2+b^2\)
Answer & Solution
Correct Answer
(B) \(a x \operatorname{sech}(t)+b y \operatorname{cosech}(t)=a^2+b^2\)
Step-by-step Solution
Detailed explanation
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…
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