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

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

Question

(D) The given line is $x - y + 1 = 0$. The curve is $x = y^2$. Let a point on the curve be $P(y^2, y)$. The distance $d$ from point $P$ to the line $x

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The given line is $x - y + 1 = 0$. The curve is $x = y^2$. Let a point on the curve be $P(y^2, y)$. The distance $d$ from point $P$ to the line $x - y + 1 = 0$ is given by $d = \frac{|y^2 - y + 1|}{\sqrt{1^2 + (-1)^2}} = \frac{|y^2 - y + 1|}{\sqrt{2}}$. Since $y^2 - y + 1 > 0$ for all real $y$,we have $d = \frac{y^2 - y + 1}{\sqrt{2}}$. To find the shortest distance,we minimize $f(y) = y^2 - y + 1$. Taking the derivative,$f'(y) = 2y - 1$. Setting $f'(y) = 0$ gives $y = \frac{1}{2}$. The minimum value is $f(\frac{1}{2}) = (\frac{1}{2})^2 - \frac{1}{2} + 1 = \frac{1}{4} - \frac{1}{2} + 1 = \frac{3}{4}$. Thus,the shortest distance is $d = \frac{3/4}{\sqrt{2}} = \frac{3}{4 \sqrt{2}} = \frac{3 \sqrt{2}}{8}$.

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

Similar Questions

Consider the conic $C: 25(x - 1)^2 + 25(y + 1)^2 = (3x - 4y)^2$. If the curve $E$ is the locus of the point of intersection of perpendicular tangents to the conic $C$,then the minimum distance between the curve $E$ and the point $(2, -1)$ is:

If $P$ is a point on the parabola $y^2=8x$ and $A$ is the point $(1,0)$,then the locus of the mid-point of the line segment $AP$ is

The equation of the parabola whose vertex and focus lie on the $x$-axis at distances $a$ and $a'$ from the origin,respectively,is

Statement $- 1 :$ For all non-zero values of $m$,$y = mx - 1/m$ is always a tangent to the parabola $y^2 = -4x$.
Statement $- 2 :$ Every tangent to the parabola $y^2 = -4x$ touches its axis at a point whose $x$-coordinate is non-negative.

If the normal at point $P(t)$ to the parabola $y^2 = 16x$ meets it again at point $Q(36, -24)$,then the maximum possible focal distance of point $P$ is-

Vedclass Test Series

Exam Paper Generator

Online Exam Module