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Question: Find the shortest distance between the parabola , \({y^2} = 4x\) and circle \({x^2} + {y^2} - 24y + ...

Find the shortest distance between the parabola , y2=4x{y^2} = 4x and circle x2+y2−24y+128=0{x^2} + {y^2} - 24y + 128 = 0 ?

Explanation

Solution

For solving this particular question, we have to find the point on the parabola and on the circle then we have to find the distance between the points by using distance formula . we need to differentiate the given function with respect to the independent variable , then equate the required result to zero. By Solving the equation, we get the values for independent variables.

Complete step by step solution:
We have to find the shortest distance between the parabola , y2=4x{y^2} = 4x and circle x2+y2−24y+128=0{x^2} + {y^2} - 24y + 128 = 0 ,
Now, consider the parabola ,
y2=4x{y^2} = 4x
Here aa is equal to one.
Therefore , we have parametric points as (t2,2t)({t^2},2t) .
Now, consider the circle ,
x2+y2−24y+128=0{x^2} + {y^2} - 24y + 128 = 0
Now subtract 128128from both the side ,
x2+y2−24y=−128{x^2} + {y^2} - 24y = - 128
Add 144144 both the side ,
(x)2+y2−24y+144=−128+144 (x)2+(y−12)2=−128+144 (x)2+(y−12)2=16 (x)2+(y−12)2=42  {(x)^2} + {y^2} - 24y + 144 = - 128 + 144 \\\ {(x)^2} + {(y - 12)^2} = - 128 + 144 \\\ {(x)^2} + {(y - 12)^2} = 16 \\\ {(x)^2} + {(y - 12)^2} = {4^2} \\\
From this we can say that ,
The centre of the circle is (0,12)(0,12) and
Radius of the circle is 44.
Now, calculate the distance between the two points that are (t2,2t)({t^2},2t) and (0,12)(0,12) is ,
=(t2−0)2+(2t−12)2 =t4+4t2−48t+144  = \sqrt {{{({t^2} - 0)}^2} + {{(2t - 12)}^2}} \\\ = \sqrt {{t^4} + 4{t^2} - 48t + 144} \\\
This distance is minimum when t4+4t2−48t+144{t^4} + 4{t^2} - 48t + 144 is minimum.
Therefore, differentiate the above expression, we will get,
4t3+8t−48=04{t^3} + 8t - 48 = 0 or
t3+2t−12=0{t^3} + 2t - 12 = 0
Therefore, we get three points that are,
(0,0),(14−213,−2+213),(14+213,−2−213)(0,0),(14 - 2\sqrt {13} , - 2 + 2\sqrt {13} ),(14 + 2\sqrt {13} , - 2 - 2\sqrt {13} ).

Note:
Questions similar in nature as that of above can be approached in a similar manner and we can solve it easily. A function f(x)f(x)encompasses a local maximum or relative maximum at xx equals to x0{x_0} if the graph of f(x)f(x)near x0{x_0} features a peak at x0{x_0}. A function f(x)f(x) features a local minimum or relative minimum at xx equals to x0{x_0} if the graph of f(x)f(x) near x0{x_0} encompasses a trough at x0{x_0}. (To make the excellence clear, sometimes the ‘plain’ maximum and minimum are called absolute maximum and minimum.)