The equation of a parabola is given.y=14x2−3x+18What are the coordinates of the focus of the parabola?Enter your answer in the boxes.
Accepted Solution
A:
Answer:The coordinates of the focus of the parabola is: [tex]\text{Focus}=(\dfrac{3}{28},\dfrac{125}{7})=(0.10714,17.8571)[/tex]Step-by-step explanation:We know that for any general equation of the parabola of the type: [tex](x-h)^2=4p(y-k)[/tex]The focus of the parabola is given by:Focus= (h,k+p)Here we are given a equation of the parabola as: [tex]y=14x^2-3x+18[/tex]On changing the equation to general form as follows:[tex]y=14(x^2-\dfrac{3}{14}x)+18\\\\\\y=14((x-\dfrac{3}{28})^2-(\dfrac{3}{28})^2)+18\\\\y=14(x-\dfrac{3}{28})^2-\dfrac{9}{56}+18\\\\\\y=14(x-\dfrac{3}{28})^2+\dfrac{999}{56}\\\\y-\dfrac{999}{56}=14(x-\dfrac{3}{28})^2\\\\(x-\dfrac{3}{28})^2=\dfrac{1}{14}(y-\dfrac{999}{56})\\\\(x-\dfrac{3}{28})^2=4\times \dfrac{1}{56}(y-\dfrac{999}{56})[/tex]Hence, we have:[tex]h=\dfrac{3}{28}\ ,\ k=\dfrac{999}{56}\ ,\ p=\dfrac{1}{56}[/tex]Hence,[tex]k+p=\dfrac{1000}{56}=\dfrac{125}{7}[/tex]Hence, focus is:[tex]\text{Focus}=(\dfrac{3}{28},\dfrac{125}{7})=(0.10714,17.8571)[/tex]