Algebra> Maxima and Minima...

Find the shortest distance of the point (0,c) from the parabola y=x^2 where 0

shaleen upadhyay , 14 Years ago

Grade 12

1 Answers

Jitender Singh

Ans:
Hello Student,
Please find answer to your question below

Shortest distance of the point from the curve lie along the normal.

$y = x^2$
Any point on the curve can be written as
$(\frac{t}{2}, \frac{t^{2}}4{})$
Slope of tangent:
$y' = 2x$
$y' = 2.\frac{t}{2} = t$
Equation of tangent:
$(y-\frac{t^{2}}{4}) = t(x-\frac{t}{2})$
$tx - y - \frac{t^{2}}{4} = 0$
Distance of the point (0,c) from the line ‘L’:
$|\frac{ct-\frac{t^{2}}{4}}{\sqrt{1+t^{2}}}|$ …..........(1)
Minimize the distance:
$\frac{\partial L}{\partial t} = 0$
$\frac{\sqrt{1+t^{2}}.(c-\frac{t}{2})-(ct-\frac{t^{2}}{4}).\frac{t}{\sqrt{1+t^{2}}}}{1+t^{2}} = 0$
$(1+t^{2})(c-\frac{t}{2})-t(ct-\frac{t^{2}}{4}) = 0$
$(c-\frac{t}{2}+ct^{2}-\frac{t^{3}}{2})-(ct^{2}-\frac{t^{3}}{4}) = 0$
$c-\frac{t}{2}-\frac{t^{3}}{4} = 0$
You have value of ‘t’ from here & put in (1)

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