The equation of the normal to the curve x=a c | vedclass

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(B) The given curve is $x=a \cosh(t)$ and $y=b \sinh(t)$,which represents the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. The slope of the tangent

Vedclass · Accepted Answer

$ax \operatorname{sech}(t)+by \operatorname{cosech}(t)=a^2+b^2$

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(B) The given curve is $x=a \cosh(t)$ and $y=b \sinh(t)$,which represents the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. The slope of the tangent at point $t$ is $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{b \cosh(t)}{a \sinh(t)}$. The slope of the normal is $-\frac{dx}{dy} = -\frac{a \sinh(t)}{b \cosh(t)}$. The equation of the normal at point $(a \cosh(t), b \sinh(t))$ is given by: $y - b \sinh(t) = -\frac{a \sinh(t)}{b \cosh(t)} (x - a \cosh(t))$. Multiplying by $b \cosh(t)$: $by \cosh(t) - b^2 \sinh(t) \cosh(t) = -ax \sinh(t) + a^2 \sinh(t) \cosh(t)$. Rearranging the terms: $ax \sinh(t) + by \cosh(t) = (a^2+b^2) \sinh(t) \cosh(t)$. Dividing both sides by $\sinh(t) \cosh(t)$: $\frac{ax}{\cosh(t)} + \frac{by}{\sinh(t)} = a^2+b^2$. This can be written as $ax \operatorname{sech}(t) + by \operatorname{cosech}(t) = a^2+b^2$.

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

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