On the parabola y = x^2,the point at the leas | vedclass

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

Question

(A) The shortest distance between a curve and a line occurs at a point where the tangent to the curve is parallel to the given line.Given the parabola

Vedclass · Accepted Answer

$(1, 1)$

Vedclass · Answer

(A) The shortest distance between a curve and a line occurs at a point where the tangent to the curve is parallel to the given line.Given the parabola $y = x^2$,the slope of the tangent at any point $(x, y)$ is given by $\frac{dy}{dx} = 2x$.The given line is $y = 2x - 4$,which has a slope of $2$.For the tangent to be parallel to the line,we set the slope of the tangent equal to the slope of the line:$2x = 2 \Rightarrow x = 1$.Substituting $x = 1$ into the parabola equation $y = x^2$,we get $y = (1)^2 = 1$.Thus,the point on the parabola closest to the line is $(1, 1)$.

On the parabola $y = x^2$,the point at the least distance from the straight line $y = 2x - 4$ is

Similar Questions

The length of the latus rectum of the parabola $20(x^2+y^2-6x-2y+10) = (4x-2y-5)^2$ is

The normal to the parabola ${y^2 = 8x}$ at the point $(2, 4)$ meets the parabola again at the point

The tangent to the parabola $y^2 = 4ax$ at the point $(a, 2a)$ makes an angle with the $x$-axis equal to

If the line $y = mx + c$ is a tangent to the parabola $y^2 = 4a(x + a)$,then $ma + \frac{a}{m}$ is equal to

The equation of the parabola with focus $(a, b)$ and directrix $\frac{x}{a} + \frac{y}{b} = 1$ is given by

Vedclass Test Series

Exam Paper Generator

Online Exam Module