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

$A = (-2, 0)$ and $P$ is a point on the parabola $y^2 = 8x$. If $Q$ bisects $\overline{AP}$ and the locus of $Q$ is a parabola,then its focus is

What is the equation of the tangent to the curve $x = at^2, y = 2at$ at any point $t$?

What does the curve defined parametrically by $x = t^2 + t + 1$ and $y = t^2 - t + 1$ represent?

If the locus of a point which divides a chord with slope $2$ of the parabola $y^2=4x$ internally in the ratio $1:3$ is a parabola,then its vertex is

The length of the subnormal to the parabola $y^2 = 4ax$ at any point is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module