A curve is represented by the equations x = s | vedclass

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

Question

(D) Given the parametric equations $x = \sec^2 t$ and $y = \cot t$.Eliminating $t$ using the identity $\sec^2 t = 1 + 	an^2 t = 1 + \frac{1}{\cot^2 t

Vedclass · Accepted Answer

$\frac{3\sqrt{5}}{2}$

Vedclass · Answer

(D) Given the parametric equations $x = \sec^2 t$ and $y = \cot t$.Eliminating $t$ using the identity $\sec^2 t = 1 + 	an^2 t = 1 + \frac{1}{\cot^2 t}$,we get $x = 1 + \frac{1}{y^2}$,which simplifies to $y^2(x - 1) = 1$.At $t = \pi / 4$,$x = \sec^2(\pi / 4) = 2$ and $y = \cot(\pi / 4) = 1$. So,$P = (2, 1)$.Differentiating $y^2(x - 1) = 1$ with respect to $x$: $2y \frac{dy}{dx}(x - 1) + y^2 = 0$.At $P(2, 1)$,$2(1) \frac{dy}{dx}(2 - 1) + 1^2 = 0 \implies 2 \frac{dy}{dx} + 1 = 0 \implies \frac{dy}{dx} = -1/2$.The equation of the tangent at $P(2, 1)$ is $y - 1 = -\frac{1}{2}(x - 2) \implies 2y - 2 = -x + 2 \implies x + 2y = 4$.Substitute $x = 4 - 2y$ into the curve equation $y^2(x - 1) = 1$: $y^2(4 - 2y - 1) = 1 \implies y^2(3 - 2y) = 1 \implies 3y^2 - 2y^3 = 1 \implies 2y^3 - 3y^2 + 1 = 0$.Since $P$ is a point of tangency,$y = 1$ is a double root. Dividing $2y^3 - 3y^2 + 1$ by $(y - 1)^2 = y^2 - 2y + 1$,we get $(y - 1)^2(2y + 1) = 0$.The other root is $y = -1/2$. Then $x = 4 - 2(-1/2) = 4 + 1 = 5$. So,$Q = (5, -1/2)$.The distance $|PQ| = \sqrt{(5 - 2)^2 + (-1/2 - 1)^2} = \sqrt{3^2 + (-3/2)^2} = \sqrt{9 + 9/4} = \sqrt{45/4} = \frac{3\sqrt{5}}{2}$.

$A$ curve is represented by the equations $x = \sec^2 t$ and $y = \cot t$,where $t$ is a parameter. If the tangent at the point $P$ on the curve where $t = \pi / 4$ meets the curve again at the point $Q$,then $|PQ|$ is equal to

Similar Questions

The number of points, at which the function $f(x) = \max\{6x, 2+3x^2\} + |x-1| |\cos(x^2 - 1/4)|, x \in (-\pi, \pi)$, is not differentiable, is ————

Suppose $a$ is a positive real number such that $a^5-a^3+a=2$. Then,

The number of points,where the curve $f(x) = e^{8x} - e^{6x} - 3e^{4x} - e^{2x} + 1$,$x \in R$ cuts the $x$-axis,is equal to

If $y = \log(\tan(x/2)) + \sin^{-1}(\cos x)$,then $dy/dx$ is

Differentiation of $(x^2-5x+8) \times (x^3+7x+9)$ can be done by

Vedclass Test Series

Exam Paper Generator

Online Exam Module