Respuesta :
Given:
Focus of a parabola = [tex]\left(1,\dfrac{1}{2}\right)[/tex]
Directrix: [tex]y=3[/tex]
To find:
The equation of the parabola.
Solution:
The equation of a vertical parabola is:
[tex]y-k=\dfrac{1}{4a}(x-h)^2[/tex] ...(A)
Where, [tex](h,k)[/tex] is center, [tex](h,k+a)[/tex] is focus and [tex]y=k-a[/tex] is the directrix.
On comparing the focus, we get
[tex](h,k+a)=\left(1,\dfrac{1}{2}\right)[/tex]
[tex]h=1[/tex]
[tex]k+a=\dfrac{1}{2}[/tex] ...(i)
On comparing the directrix, we get
[tex]k-a=3[/tex] ...(ii)
Adding (i) and (ii), we get
[tex]2k=\dfrac{7}{2}[/tex]
[tex]k=\dfrac{7}{4}[/tex]
Putting [tex]k=\dfrac{7}{4}[/tex] is (i), we get
[tex]\dfrac{7}{4}+a=\dfrac{1}{2}[/tex]
[tex]a=\dfrac{1}{2}-\dfrac{7}{4}[/tex]
[tex]a=\dfrac{-5}{4}[/tex]
Putting [tex]a=\dfrac{-5}{4},h=1,k=\dfrac{7}{4}[/tex] in (A), we get
[tex]y-\dfrac{7}{4}=\dfrac{1}{4\times \dfrac{-5}{4}}(x-1)^2[/tex]
[tex]y-\dfrac{7}{4}=\dfrac{-1}{5}(x^2-2x+1)[/tex]
[tex]y-\dfrac{7}{4}=-\dfrac{1}{5}(x^2)-\dfrac{1}{5}(-2x)-\dfrac{1}{5}(1)[/tex]
[tex]y=-\dfrac{1}{5}x^2+\dfrac{2}{5}x-\dfrac{1}{5}+\dfrac{7}{4}[/tex]
On further simplification, we get
[tex]y=-\dfrac{1}{5}x^2+\dfrac{2}{5}x+\dfrac{35-4}{20}[/tex]
[tex]y=-\dfrac{1}{5}x^2+\dfrac{2}{5}x+\dfrac{31}{20}[/tex]
Therefore, the equation of the parabola is [tex]y=-\dfrac{1}{5}x^2+\dfrac{2}{5}x+\dfrac{31}{20}[/tex].
Note: Option C is correct but the leading coefficient should be negative.
