The distance between the origin and the norma | vedclass

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

Question

(C) Given curve is $y = e^{2x} + x^2$. At $x = 0$,$y = e^0 + 0^2 = 1$. So,the point of contact is $(0, 1)$. Differentiating with respect to $x$,we get

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Given curve is $y = e^{2x} + x^2$. At $x = 0$,$y = e^0 + 0^2 = 1$. So,the point of contact is $(0, 1)$. Differentiating with respect to $x$,we get $\frac{dy}{dx} = 2e^{2x} + 2x$. At $x = 0$,the slope of the tangent $m_t = 2e^0 + 2(0) = 2$. The slope of the normal $m_n = -\frac{1}{m_t} = -\frac{1}{2}$. The equation of the normal at $(0, 1)$ is $y - 1 = -\frac{1}{2}(x - 0)$. $2y - 2 = -x$,which simplifies to $x + 2y - 2 = 0$. The distance $d$ from the origin $(0, 0)$ to the line $Ax + By + C = 0$ is given by $d = \frac{|C|}{\sqrt{A^2 + B^2}}$. Here,$A = 1$,$B = 2$,and $C = -2$. $d = \frac{|-2|}{\sqrt{1^2 + 2^2}} = \frac{2}{\sqrt{1 + 4}} = \frac{2}{\sqrt{5}}$ units.

The distance between the origin and the normal to the curve $y = e^{2x} + x^2$ drawn at $x = 0$ is . . . . . . units

Similar Questions

What is the angle between the curves $y = x^2$ and $x = y^2$ at the point $(1, 1)$?

Find the coordinates of a point on the curve $y=x^2-3x+2$,at which the tangent drawn to this curve is perpendicular to the line $y=x$.

The point on the curve $y = 2 + \sqrt{4x + 1}$ at which the slope of the tangent is $\frac{2}{5}$ is:

If $y=4x-5$ is tangent to the curve $y^2=px^3+q$ at $(2,3)$,then

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:

Vedclass Test Series

Exam Paper Generator

Online Exam Module