The shortest distance between the line y - x | vedclass

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let the point on the curve $x = y^2$ be $P(a^2, a)$.The equation of the line is $x - y + 1 = 0$.The perpendicular distance $D$ from point $P(a^2, a)$ to the line $x - y + 1 = 0$ is given by the formula $D = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$.Substituting the values,we get $D = \frac{|a^2 - a + 1|}{\sqrt{1^2 + (-1)^2}} = \frac{a^2 - a + 1}{\sqrt{2}}$ (since $a^2 - a + 1 > 0$ for all real $a$).To find the minimum distance,we minimize the expression $f(a) = a^2 - a + 1$.Completing the square,$f(a) = (a - \frac{1}{2})^2 + \frac{3}{4}$.The minimum value of $f(a)$ occurs at $a = \frac{1}{2}$,which is $\frac{3}{4}$.Therefore,the minimum distance $D_{\min} = \frac{3/4}{\sqrt{2}} = \frac{3}{4\sqrt{2}} = \frac{3\sqrt{2}}{8}$.

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

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