The shortest distance between the line y = x | vedclass

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

Question

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

Vedclass · Accepted Answer

$\frac{7}{4\sqrt{2}}$

Vedclass · Answer

(C) The shortest distance between a line and a curve occurs at a point on the curve where the tangent is parallel to the given line.The given line is $y = x$,which has a slope of $m = 1$.The curve is $y^2 = x - 2$. Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 1$,so $\frac{dy}{dx} = \frac{1}{2y}$.Setting the slope of the tangent equal to the slope of the line: $\frac{1}{2y} = 1 \Rightarrow y = \frac{1}{2}$.Substituting $y = \frac{1}{2}$ into the curve equation: $(\frac{1}{2})^2 = x - 2 \Rightarrow \frac{1}{4} = x - 2 \Rightarrow x = 2 + \frac{1}{4} = \frac{9}{4}$.So,the point on the curve is $P(\frac{9}{4}, \frac{1}{2})$.The shortest distance is the perpendicular distance from point $P(\frac{9}{4}, \frac{1}{2})$ to the line $x - y = 0$.Distance $d = \frac{|x_1 - y_1|}{\sqrt{1^2 + (-1)^2}} = \frac{|\frac{9}{4} - \frac{1}{2}|}{\sqrt{2}} = \frac{|\frac{9-2}{4}|}{\sqrt{2}} = \frac{7}{4\sqrt{2}}$.

The shortest distance between the line $y = x$ and the curve $y^2 = x - 2$ is

Similar Questions

Two tangents are drawn from a point $(-2, -1)$ to the curve $y^2 = 4x$. If $\alpha$ is the angle between them,then $|\tan \alpha|$ is equal to:

The focus of the parabola $y^2 - x - 2y + 2 = 0$ is

$A$ tangent to the curve $x = a t^{2}, y = 2 a t$ is perpendicular to the $X$-axis. Then the point of contact is:

What is the equation of the directrix of the parabola $x^2 = -8y$?

The distance of the focus of the parabola $y^2 = 16x$ from its directrix is...

Vedclass Test Series

Exam Paper Generator

Online Exam Module