What is the slope of the normal to the parabo | vedclass

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

Question

(B) Given the equation of the parabola is $x^2 + 4y = 0$.We can rewrite this as $4y = -x^2$,which implies $y = -\frac{1}{4}x^2$.To find the slope of t

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) Given the equation of the parabola is $x^2 + 4y = 0$.We can rewrite this as $4y = -x^2$,which implies $y = -\frac{1}{4}x^2$.To find the slope of the tangent at any point $(x, y)$,we differentiate with respect to $x$:$\frac{dy}{dx} = -\frac{1}{4} 	imes 2x = -\frac{x}{2}$.At the point $(2, -1)$,the slope of the tangent $(m_t)$ is:$m_t = -\frac{2}{2} = -1$.The slope of the normal $(m_n)$ is the negative reciprocal of the slope of the tangent:$m_n = -\frac{1}{m_t} = -\frac{1}{-1} = 1$.Therefore,the slope of the normal is $1$.

What is the slope of the normal to the parabola $x^2 + 4y = 0$ at the point $(2, -1)$?

Similar Questions

The position of a moving point in the $XY$-plane at time $t$ is given by $\left( (u \cos \alpha)t, (u \sin \alpha)t - \frac{1}{2}gt^2 \right)$,where $u, \alpha, g$ are constants. The locus of the moving point is

What is the length of the chord of contact of the tangents drawn from the point $(-2, -1)$ to the parabola $y^2 = 4x$?

If the double ordinate of the parabola $y^2 = 8x$ is of length $16$,then the angle subtended by it at the vertex of the parabola is

If $(2,3)$ is the vertex and $(3,2)$ is the focus of a parabola,then its equation is

If $x = my + c$ is a normal to the parabola ${x^2} = 4ay$,then the value of $c$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module