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

(D) The given line is $x - y + 1 = 0$,which has a slope of $1$.For the shortest distance,the tangent to the curve $x = y^2$ must be parallel to the gi

Vedclass · Accepted Answer

$\frac{3\sqrt{2}}{8}$

Vedclass · Answer

(D) The given line is $x - y + 1 = 0$,which has a slope of $1$.For the shortest distance,the tangent to the curve $x = y^2$ must be parallel to the given line.Differentiating $x = y^2$ with respect to $y$,we get $\frac{dx}{dy} = 2y$.Since the slope of the tangent is $\frac{dy}{dx} = 1$,we have $\frac{1}{2y} = 1$,which gives $y = \frac{1}{2}$.Substituting $y = \frac{1}{2}$ into $x = y^2$,we get $x = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$.So,the point $P$ on the curve is $\left(\frac{1}{4}, \frac{1}{2}\right)$.The shortest distance is the perpendicular distance from point $P\left(\frac{1}{4}, \frac{1}{2}\right)$ to the line $x - y + 1 = 0$.Distance $= \frac{|\frac{1}{4} - \frac{1}{2} + 1|}{\sqrt{1^2 + (-1)^2}} = \frac{|\frac{3}{4}|}{\sqrt{2}} = \frac{3}{4\sqrt{2}} = \frac{3\sqrt{2}}{8}$.

The shortest distance between the line $y - x = 1$ and the curve $x = y^2$ is

Similar Questions

If the lines $y = x$ and $y = -x$ intersect the parabola $y^2 = 4x$ at points $A$ and $B$ respectively,other than the origin,what is the length of $AB$?

The length of the latus rectum of $3x^{2} - 4y + 6x - 3 = 0$ is

For the parabola $y^2 + 8x - 12y + 20 = 0$,which of the following is $NOT$ true?

If $x^2 = 8ay$ is the transformed equation of $x^2 - 4y + 6x + 15 = 0$ when the origin is shifted to the point $(\alpha, \beta)$ by translation of axes,then $2\alpha + 8\beta^2 =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module