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

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

Grade:12

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
9 years ago
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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