Show that the tangent at any point $ heta$ on the curve $x = c\sec heta$, $y = c an heta$ is $y\sin heta = x - c\cos heta$.

Question

Show that the tangent at any point $	heta$ on the curve $x = c\sec	heta$, $y = c	an	heta$ is $y\sin	heta = x - c\cos	heta$.

Accepted Answer

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> [!TLDR]
> Find $dy/dx$ from the parametric derivatives, write the point-slope tangent equation, and simplify using $1-\sin^2	heta=\cos^2	heta$.
>
> Given $x=c\sec	heta$, $y=c	an	heta$.
>
> $$\frac{dx}{d	heta} = c\sec	heta	an	heta, \qquad \frac{dy}{d	heta} = c\sec^{2}	heta$$
>
> $$\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{c\sec^{2}	heta}{c\sec	heta	an	heta} = \frac{\sec	heta}{	an	heta} = \frac{1}{\sin	heta}$$
>
> The tangent line at the point $(c\sec	heta, c	an	heta)$ has equation:
> $$y - c	an	heta = \frac{1}{\sin	heta}(x - c\sec	heta)$$
>
> Multiply both sides by $\sin	heta$:
> $$y\sin	heta - c	an	heta\sin	heta = x - c\sec	heta$$
>
> Note $	an	heta\sin	heta = \dfrac{\sin^{2}	heta}{\cos	heta}$ and $\sec	heta=\dfrac{1}{\cos	heta}$, so: …

Q.Show that the tangent at any point $\theta$ on the curve $x = c\sec\theta$ , $y = c\tan\theta$ is $y\sin\theta = x - c\cos\theta$ .

Concept understanding — Tangents and Normals (Lengths)

Tangents and Normals to a Curve

Q.Show that the tangent at any point θ\thetaθ on the curve x=csec⁡θx = c\sec\thetax=csecθ, y=ctan⁡θy = c\tan\thetay=ctanθ is ysin⁡θ=x−ccos⁡θy\sin\theta = x - c\cos\thetaysinθ=x−ccosθ.

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Q.Show that the tangent at any point $\theta$ on the curve $x = c\sec\theta$ , $y = c\tan\theta$ is $y\sin\theta = x - c\cos\theta$ .

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