Sure, let's solve for the coordinates of point [tex]\( H \)[/tex], given that it lies at [tex]\( (x, 9) \)[/tex] and the distance between [tex]\( F \)[/tex] and [tex]\( H \)[/tex] is 10 units.
First, let's denote the coordinates of point [tex]\( F \)[/tex] as [tex]\( (f_x, f_y) \)[/tex]. Given that [tex]\( F \)[/tex] lies on the x-axis, we can assume [tex]\( f_y = 0 \)[/tex]. So the coordinates of point [tex]\( F \)[/tex] are [tex]\( (f_x, 0) \)[/tex].
Next, we'll use the distance formula to find the possible values of [tex]\( x \)[/tex]. The distance formula between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is:
[tex]\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
In this case, the distance between [tex]\( F \)[/tex] and [tex]\( H \)[/tex] is given as 10 units. Using the coordinates:
- [tex]\( F(f_x, 0) \)[/tex]
- [tex]\( H(x, 9) \)[/tex]
The distance formula becomes:
[tex]\[ \sqrt{(x - f_x)^2 + (9 - 0)^2} = 10 \][/tex]
Simplify this equation:
[tex]\[ \sqrt{(x - f_x)^2 + 81} = 10 \][/tex]
Square both sides to eliminate the square root:
[tex]\[ (x - f_x)^2 + 81 = 100 \][/tex]
Subtract 81 from both sides:
[tex]\[ (x - f_x)^2 = 19 \][/tex]
Take the square root of both sides:
[tex]\[ x - f_x = \pm \sqrt{19} \][/tex]
This gives us two possible solutions:
[tex]\[ x - f_x = \sqrt{19} \][/tex]
[tex]\[ x = f_x + \sqrt{19} \][/tex]
or
[tex]\[ x - f_x = -\sqrt{19} \][/tex]
[tex]\[ x = f_x - \sqrt{19} \][/tex]
Therefore, the two possible values of [tex]\( x \)[/tex] are:
[tex]\[ x = f_x + \sqrt{19} \][/tex]
and
[tex]\[ x = f_x - \sqrt{19} \][/tex]
So, the coordinates of point [tex]\( H \)[/tex] can be either [tex]\( (f_x + \sqrt{19}, 9) \)[/tex] or [tex]\( (f_x - \sqrt{19}, 9) \)[/tex].
