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

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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 $d = \frac{3/4}{\sqrt{2}} = \frac{3}{4 \sqrt{2}} = \frac{3 \sqrt{2}}{8}$.

List-$A$	List-$B$
$(A)$. The vertex of the parabola $y^2+4x-2y+3=0$ is	$(I)$. $\left(\frac{5}{4}, 1\right)$
$(B)$. The vertex of the parabola $x^2+8x+12y+4=0$ is	$(II)$. $\left(1, \frac{5}{4}\right)$
$(C)$. The focus of the parabola $y^2-x-2y+2=0$ is	$(III)$. $\left(-\frac{1}{2}, 1\right)$
$(D)$. The focus of the parabola $x^2-2x-8y-23=0$ is	$(IV)$. $(1, -1)$
	$(V)$. $(-4, 1)$

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

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