Question

Find the minimum distance between the curve `y^2 = 4ax` & `x^2 + y^2 -12x + 31=0`?

The vale of a is 1.

Answer 1

The points are `(4,-4)` and `(4,4)` and distance is `sqrt5`

Explanation:

As `a=1`, the equation of parabola is `y^2=4x`. The parametric form of its equation would be `(t^2,2t)`.

Now, `x^2+y^2-12x+31=0` is a circle, whose center is `(6,0)` and radius is `sqrt(6^2-31)=sqrt5`.

The distance between `(t^2,2t)` and `(6,0)` is

`sqrt((t^2-6)^2+(2t)^2)`

or `sqrt(t^4-8t^2+36)`

This distance willbe minimum when `t^4-8t^2+36` is minimum i.e. its first derivative`4t^3-16t=0` or `t^3-4t=0` i.e. `t(t-2)(t+2)=0`.

i.e. `t=0,2` or `-2` and points corresponding to these are

`(0,0)`, `(4,-4)` and `(4,4)`

Of these while distance of `(4,-4)` and `(4,4)` is `sqrt5` that of `(0,0)` is `6-sqrt5`.

Another way could be to find a normal to parabola which passes through the center of circle `(6,0)`.

As parametric form of equation of parabola is `(t^2,2t)`, slope of tangent is given by `(((dy)/(dt))/((dx)/(dt)))=2/(2t)=1/t`. Hence slope of normal is `-t` and equation of normal is `y-2t=-t(x-t^2)` or `y+xt=t^3+2t`.

If this passes through `(6,0)`, we have `6t=t^3+2t` or `t^3-4t=0` again giving us same solution.

Hence the points are `(4,-4)` and `(4,4)` and distance is `sqrt5`
graph{(y^2-4x)(x^2+y^2-12x+31)=0 [-10, 10, -5, 5]}

Answer 2

Given equations of conics are
* Parabola `y^2=4ax`,where `a=1`
`or,y^2=4x......[1]`
* Circle `x^2+y^2-12x+31=0`,which can be written instandard form as
`(x-6)^2+y^2=(sqrt5)^2..........[2]`.
So coordinates of the center of the circle `(6,0)` and it's radius`=sqrt5`

Now to find out the minimum distance between them we can apply our simple geometric logic that in approaching towards a circle from any point outside circle, the distance covered will be minimum if approach remains directed towards its center.

Hence distance (D) of any point (para-metrically `(t^2,2t)`) on the parabola to the circle (while approaching towards its center
`(6.0)` will be given by

`D=sqrt((t^2-6)^2+(2t-0)^2)-sqrt5`

`=>D=sqrt(t^4-12t^2+36+4t^2)-sqrt5`

`=>D=sqrt(t^4-2*t^2*4+4^2+20)-sqrt5`

`=>D=sqrt((t^2-4)^2+20)-sqrt5`

`=>D_"min"=sqrt20-sqrt5`, for `t=pm2`

So `D_"min"=2sqrt5-sqrt5=sqrt5`

Alternative Metod .

Minimum distance between the conics is the minimum length of the line segment in between them on the common normal of the two conics.
The coordinates of any point on the given parabola `y^2=4x` is `(t^2,2t)`. Slope of the tangent to the parabola at this point will be

`[(dy)/(dx)] _ ("("t^2,2t")")= [2/y]_("("t^2,2t")")=1/t`

So slope of the normal will be `m=-t`

The equation of the normal will be

`y-2t=-t(x-t^2)`

Every normal of a circle passes through its center. So the equation of common normal should pass through the center of the circle.

Hence we can find the possible values of `t` by inserting (x=6 and y=0) in the above equation of the normal

`0-2t=-t(6-t^2)`

`=>t^3-6t+2t=0`

`=>t(t^2-4)=0`

`=>t(t+2)(t-2)=0`

So we get three values of `t=0,-2,+2` .that means there exist three common normals,passing through three different points of contact `O(0,0);C(4,4);E(4,-4)` on the parabola.

The equations of common normals are

1) Passing through `(0,0)`

`y-2*0=-0*(x-0)`
`=>y=0.......[3]`

2) Passing through `(4,4)`

`y-2*2=-2*(x-2^2)`
`=>y=-2x+12........[4]`

3) Passing through `(4,-4)`

`y-2(-2)=-(-2)*(x-(-2)^2)`
`=>y=2x-12.........[5]`

Solving [2] and [3] we get `x=6pmsqrt5 and y=0`

So length of the line segment on common normal passing through `(0,0)` in between two conics will be `OA=6-sqrt5`

Solving [2] and [4]

`(x-6)^2+(12-2x)^2=5`

`=>5x^2-60x+175=0`

`=>x^2-12x+35=0`

`=>(x-5)(x-7)=0`

We get `x=5and7`
and corresponding values of y coordinates are `y=-2 and 2`

So the common normal passing through C(4,-4) will intersect the circle at `D(5,2) and G(7,-2)`

Here the distance between the conics along [4]

`CD=sqrt((5-4)^2+(2-4)^2=sqrt5`

Solving [2] and [5]

`(x-6)^2+(2x-12)^2=5`

`=>5x^2-60x+175=0`

`=>x^2-12x+35=0`
`=>(x-5)(x-7)=0`
We get `x=5and7`
and corresponding values of y-coordinates are `y=-2 and 2`
So the common normal passing through E(4,-4) will intersect the circle at `F(5,-2) and H(7,2)`
Here the distance between the conics along [5]

`EF=sqrt((5-4)^2+(-2+4)^2=sqrt5`

Hence minimum distance between two conic is `sqrt5`

Answer 3

Using the so called Lagrange Multipliers

Calling

`g_1(x_1,y_1) = y_1^2-4x_1`
`g_2(x_2,y_2) = x_2^2+y_2^2-12x_2+31`

and

`d^2(x_1,y_1,x_2,y_2) = (x_1-x_2)^2+(y_1-y_2)^2`

then

`L(x_1,y_1,x_2,y_2,lambda_1,lambda_2) = d^2(x_1,y_1,x_2,y_2)+lambda_1 g_1(x_1,y_1)+lambda_2 g_2(x_2,y_2)`

The stationary conditions are

#{ (L_(x_1)=2 (x_1 - x_2)-4 lambda_1 =0), ( L_(x_2)=lambda_2 ( 2 x_2-12)-2 (x_1 - x_2) =0), (L_(y_1)=2 lambda_1 y_1 + 2 (y_1 - y_2)=0), (L_(y_2)=2 lambda_2 y_2-2 (y_1 - y_2) =0), ( L_(lambda_1)=y_1^2-4 x_1=0), (L_(lambda_2)=31 - 12 x_2 + x_2^2 + y_2^2=0):}#

Solving for `x_1,y_1,x_2,y_2,lambda_1,lambda_2` we obtain

#((x_1,x_2,y_1,y_2,lambda_1,lambda_2,d),(4, 5, -4, -2, -1/2, 1,sqrt5),(4, 5,4, 2, -1/2 ,1,sqrt5),(4, 7, -4, 2, -3/2, -3,3sqrt5),(4,7, 4, -2, -3/2, -3,3sqrt5),( 0, 6 - sqrt[5], 0, 0, 1/2 (sqrt[5]-6), 1/5 (31 - 6 (6 - sqrt[5])),6-sqrt5),( 0, 6 + sqrt[5], 0, 0, -1/2 (6 + sqrt[5]), 1/5 (31 - 6 (6 + sqrt[5])),6+sqrt5))#

and the minimum is `sqrt5`