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Question

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


A
832
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B
43
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C
34
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D
328
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Solution

The correct option is A $$\dfrac{8}{3\sqrt{2}}$$
$$ y^2 = x \\ \Rightarrow 2y = \dfrac{dx}{dy} \, [ \textrm{Differentiating both sides w.r.t. 'y' } ] \\ \left[ - \dfrac{dx}{dy} \right]_{x_1, y_1} = -2y_1 $$
$$ \therefore $$ Slope of the normal of the parabola $$y^x$$ at $$P ( x_1, y_1) = -2y_1 $$
Again slope of the line y - x = 1 is 1.
$$ \therefore (-2y_1).(1) = -1 \\ \Rightarrow y_1 = \dfrac{1}{2} $$
Again $$P (x_1, y_1) $$ lie on $$ y^2 = x $$
$$ \therefore x_1 = y_1^2 = \dfrac{1}{4}. $$
$$ \therefore $$ The point P is $$ \left( \dfrac{1}{4} , \dfrac{1}{2} \right) $$
$$ \therefore $$ The distance from $$ \left( \dfrac{1}{4} , \dfrac{1}{2} \right) $$ to y - x = 1
$$ d = \dfrac{| \frac{1}{2} - \frac{1}{4} - 1 | } { \sqrt{ (1)^2 + (-1) ^2 }} = \dfrac{ |-\frac{1}{2} - \frac{1}{4} | } {\sqrt2} \\ = \dfrac{3}{4\sqrt2} = \dfrac{3\sqrt2}{8} $$


1880956_1331058_ans_73ec6b1fe7a545f1b6ed4ce6fb16e48d.png

Mathematics

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