The equation of a normal to the curve x=4 sec | vedclass

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(D) Given the parametric equations of the curve: $x=4 \sec 	heta$ and $y=4 	an^2 	heta$. First,find the derivatives with respect to $	heta$: $\fra

Vedclass · Accepted Answer

$x+2 \sqrt{2} y=12 \sqrt{2}$

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(D) Given the parametric equations of the curve: $x=4 \sec 	heta$ and $y=4 	an^2 	heta$. First,find the derivatives with respect to $	heta$: $\frac{dx}{d	heta} = 4 \sec 	heta 	an 	heta$ and $\frac{dy}{d	heta} = 8 	an 	heta \sec^2 	heta$. The slope of the tangent is $\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{8 	an 	heta \sec^2 	heta}{4 \sec 	heta 	an 	heta} = 2 \sec 	heta$. At $	heta = \frac{\pi}{4}$,the slope of the tangent is $2 \sec(\frac{\pi}{4}) = 2 \sqrt{2}$. The slope of the normal is $-\frac{1}{	ext{slope of tangent}} = -\frac{1}{2 \sqrt{2}}$. At $	heta = \frac{\pi}{4}$,the point $(x, y)$ is $(4 \sec \frac{\pi}{4}, 4 	an^2 \frac{\pi}{4}) = (4 \sqrt{2}, 4)$. The equation of the normal is $y - y_1 = m_n(x - x_1)$: $y - 4 = -\frac{1}{2 \sqrt{2}}(x - 4 \sqrt{2})$. Multiplying by $2 \sqrt{2}$: $2 \sqrt{2} y - 8 \sqrt{2} = -x + 4 \sqrt{2}$. Rearranging gives $x + 2 \sqrt{2} y = 12 \sqrt{2}$.

The equation of a normal to the curve $x=4 \sec \theta$ and $y=4 \tan^2 \theta$ at $\theta=\frac{\pi}{4}$ is

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