The equation for the distance between any point and the point (0, 9) is:
[tex]d=\sqrt{x^2+(y-9)^2}[/tex]The curve given in the problem is:
[tex]y=x^2+5[/tex]If we solve for 'x²', we can substitute in the distance equation and obtain a function for the distance of any point in the curve to (0, 9):
[tex]x^2=y-5[/tex]And substitute:
[tex]d(y)=\sqrt{(y-5)+(x-9)^2}[/tex]SImplify:
[tex]d(y)=\sqrt{y^2-17y+76}[/tex]And now, we need to calculate the first and second derivatives:
[tex]d^{\prime}(x)=\frac{2y-17}{2\sqrt{y^2-17y+76}}[/tex][tex]d^{\prime}^{\prime}(y)=\frac{15}{4(y^2-17y+76)^{\frac{3}{2}}}[/tex]Then, find the critical points of d(y). Since this function is a quotient, it will be equal to 0 when the numerator is equal to 0:
[tex]\begin{gathered} 2y-17=0 \\ . \\ y=\frac{17}{2} \end{gathered}[/tex]Now, evaluate this value into the second derivative:
[tex]d^{\prime}^{\prime}(\frac{17}{2})=\frac{15}{4((\frac{17}{2})^2-17\cdot\frac{17}{2}+76)^{\frac{3}{2}}}\approx0.516546[/tex]Since is a positive value, the function d(y) has a minimum in y = 17/2
Next, we need to find the values of x. We use the equation of the curve:
[tex]\begin{gathered} \frac{17}{2}=x^2+5 \\ . \\ x^2=\frac{17}{2}-5 \\ . \\ x=\pm\sqrt{\frac{7}{2}}=\pm\frac{\sqrt{14}}{2} \end{gathered}[/tex]Thus, the point on the curve closest to (0, 9), rounded to two decimals, are:
[tex](-1.87,8.5)[/tex][tex](1.87,8.5)[/tex]
