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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

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

Vedclass · Answer

(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

Similar Questions

Length of the subtangent at $(x_{1}, y_{1})$ on $x^{n} y^{m}=a^{m+n}, m, n>0$,is

Consider a curve $y=y(x)$ in the first quadrant as shown in the figure. Let the area $A_{1}$ be twice the area $A_{2}$. Then the normal to the curve perpendicular to the line $2x - 12y = 15$ does $NOT$ pass through the point.

The length of the segment of the tangent line to the curve $x=a \cos ^3 t, y=a \sin ^3 t$, at any point on the curve cut off by the coordinate axes is

Let $a$ be a fixed positive real number and $n$ be an arbitrary constant. For the curve $y = \frac{x^n}{a^{n-1}}$, if the length of the subnormal at any point $(\alpha, \beta)$ is proportional to $a^2$, then $n =$

If the tangent at the point $(x_{1}, y_{1})$ on the curve $y=x^{3}+3x^{2}+5$ passes through the origin,then $(x_{1}, y_{1})$ does $NOT$ lie on which of the following curves?

Vedclass Test Series

Exam Paper Generator

Online Exam Module