Question

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

Answer 1

Hint:
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 , ${y^2} = 4x$ and circle ${x^2} + {y^2} - 24y + 128 = 0$ ,
Now, consider the parabola ,
${y^2} = 4x$
Here $a$ is equal to one.
Therefore , we have parametric points as $({t^2},2t)$ .
Now, consider the circle ,
${x^2} + {y^2} - 24y + 128 = 0$
Now subtract $128$from both the side ,
${x^2} + {y^2} - 24y = - 128$
Add $144$ both the side ,
$
  {(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 $(0,12)$ and
Radius of the circle is $4$.
Now, calculate the distance between the two points that are $({t^2},2t)$ and $(0,12)$ is ,
$
   = \sqrt {{{({t^2} - 0)}^2} + {{(2t - 12)}^2}} \\
   = \sqrt {{t^4} + 4{t^2} - 48t + 144} \\
 $
This distance is minimum when ${t^4} + 4{t^2} - 48t + 144$ is minimum.
Therefore, differentiate the above expression, we will get,
$4{t^3} + 8t - 48 = 0$ or
${t^3} + 2t - 12 = 0$
Therefore, we get three points that are,
$(0,0),(14 - 2\sqrt {13} , - 2 + 2\sqrt {13} ),(14 + 2\sqrt {13} , - 2 - 2\sqrt {13} )$.


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