The shortest distance between the line x-y=1 | vedclass

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

Question

(B) The shortest distance between a curve and a line occurs at a point $P(x_0, y_0)$ on the curve where the tangent is parallel to the given line.The

Vedclass · Accepted Answer

$\frac{1}{2\sqrt{2}}$

Vedclass · Answer

(B) The shortest distance between a curve and a line occurs at a point $P(x_0, y_0)$ on the curve where the tangent is parallel to the given line.The equation of the line is $x-y=1$,which can be written as $y=x-1$. The slope of this line is $m=1$.The equation of the curve is $x^2=2y$,which implies $y=\frac{x^2}{2}$.The slope of the tangent to the curve at any point $(x_0, y_0)$ is given by $\frac{dy}{dx} = \frac{d}{dx}(\frac{x^2}{2}) = x$.Setting the slope of the tangent equal to the slope of the line,we get $x_0 = 1$.Substituting $x_0=1$ into the curve equation $y_0 = \frac{x_0^2}{2}$,we get $y_0 = \frac{1^2}{2} = \frac{1}{2}$.Thus,the point on the curve is $P(1, \frac{1}{2})$.The shortest distance from a point $(x_1, y_1)$ to the line $Ax+By+C=0$ is given by $d = \frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$.Here,the line is $x-y-1=0$,so $A=1, B=-1, C=-1$. The point is $(1, \frac{1}{2})$.Shortest distance $= \left|\frac{1(1) + (-1)(\frac{1}{2}) - 1}{\sqrt{1^2+(-1)^2}}\right| = \left|\frac{1 - \frac{1}{2} - 1}{\sqrt{2}}\right| = \left|\frac{-\frac{1}{2}}{\sqrt{2}}\right| = \frac{1}{2\sqrt{2}}$.

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

Similar Questions

If the line $7y - 4x = 10$ is a tangent to the parabola $y^2 = 4x$,find the point of contact.

$P$ is a point on the parabola $y^2 = 4ax$ $(a > 0)$ whose vertex is $A$. $PA$ is produced to meet the directrix in $D$ and $M$ is the foot of the perpendicular from $P$ on the directrix. If a circle is described on $MD$ as a diameter,then it intersects the $x$-axis at a point whose coordinates are:

$A$ circle of radius $2$ unit passes through the vertex and the focus of the parabola $y^{2}=2x$ and touches the parabola $y=\left(x-\frac{1}{4}\right)^{2}+\alpha$,where $\alpha>0$. Then $(4\alpha-8)^{2}$ is equal to

The sum of squares of all possible values of $k$,for which the area of the region bounded by the parabolas $2y^2 = kx$ and $ky^2 = 2(y - x)$ is maximum,is equal to:

Two tangent lines $l_{1}$ and $l_{2}$ are drawn from the point $(2,0)$ to the parabola $2y^{2} = -x$. If the lines $l_{1}$ and $l_{2}$ are also tangent to the circle $(x-5)^{2} + y^{2} = r$,then $17r$ is equal to.

Vedclass Test Series

Exam Paper Generator

Online Exam Module