If the normal to the curve y = x + frac{2}{x} | vedclass

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

Question

(C) Given the curve $y = x + \frac{2}{x}$.At $x = 2$,$y = 2 + \frac{2}{2} = 3$. So,the point is $(2, 3)$.Differentiating with respect to $x$,we get $\

Vedclass · Accepted Answer

$\frac{7\sqrt{5}}{2}$

Vedclass · Answer

(C) Given the curve $y = x + \frac{2}{x}$.At $x = 2$,$y = 2 + \frac{2}{2} = 3$. So,the point is $(2, 3)$.Differentiating with respect to $x$,we get $\frac{dy}{dx} = 1 - \frac{2}{x^2}$.At $x = 2$,the slope of the tangent $m_t = 1 - \frac{2}{4} = 1 - \frac{1}{2} = \frac{1}{2}$.The slope of the normal $m_n = -\frac{1}{m_t} = -2$.The equation of the normal at $(2, 3)$ is $y - 3 = -2(x - 2)$,which simplifies to $y - 3 = -2x + 4$,or $2x + y = 7$.To find the points $A$ and $B$ where the normal meets the coordinate axes:For the $x$-intercept $(y = 0)$,$2x = 7 \implies x = \frac{7}{2}$. So,$A = (\frac{7}{2}, 0)$.For the $y$-intercept $(x = 0)$,$y = 7$. So,$B = (0, 7)$.The length of $AB = \sqrt{(\frac{7}{2} - 0)^2 + (0 - 7)^2} = \sqrt{\frac{49}{4} + 49} = \sqrt{49(\frac{1}{4} + 1)} = 7 \sqrt{\frac{5}{4}} = \frac{7\sqrt{5}}{2}$.

If the normal to the curve $y = x + \frac{2}{x}$ at the point where the abscissa is $2$ meets the coordinate axes at points $A$ and $B$,find the length of $AB$.

Similar Questions

If $\beta$ is an angle between the normals drawn to the curve $x^2+3y^2=9$ at the points $P(3 \cos \theta, \sqrt{3} \sin \theta)$ and $Q(-3 \sin \theta, \sqrt{3} \cos \theta)$,where $\theta \in (0, \frac{\pi}{2})$,then:

Prove that the curves $xy=4$ and $x^{2}+y^{2}=8$ touch each other.

If the angle between the curves $y^2=4x$ and $y=e^{-x/2}$ is $\theta$,then $\operatorname{cosec}^2(\theta/2)=$

What is the equation of the tangent to the curve $x = a \cos^3 t, y = a \sin^3 t$ at point $t$?

Find the slope of the tangent to the curve $y = x^{3} - x$ at $x = 2$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module