Respuesta :
Given:
Focus of the parabola = [tex]\left(-\dfrac{1}{4},-\dfrac{2}{3}\right)[/tex]
Directrix of the parabola is [tex]y=\dfrac{3}{4}[/tex].
To find:
The equation of the parabola.
Solution:
The equation of the parabola is:
[tex](x-h)^2=4p(y-k)[/tex] ...(i)
Where, (h,k) is vertex, (h,k+p) is focus and [tex]y=k-p[/tex] is the directrix.
It is given that the focus of the parabola is at [tex][/tex].
[tex](h,k+p)=\left(-\dfrac{1}{4},-\dfrac{2}{3}\right)[/tex]
On comparing both sides, we get
[tex]h=-\dfrac{1}{4}[/tex]
[tex]k+p=-\dfrac{2}{3}[/tex] ...(ii)
Directrix of the parabola is [tex]y=\dfrac{3}{4}[/tex]. So,
[tex]k-p=\dfrac{3}{4}[/tex] ...(iii)
Adding (ii) and (iii), we get
[tex]2k=-\dfrac{2}{3}+\dfrac{3}{4}[/tex]
[tex]2k=\dfrac{-8+9}{12}[/tex]
[tex]k=\dfrac{1}{12\times 2}[/tex]
[tex]k=\dfrac{1}{24}[/tex]
Putting [tex]k=\dfrac{1}{24}[/tex] in (ii), we get
[tex]\dfrac{1}{24}+p=-\dfrac{2}{3}[/tex]
[tex]p=-\dfrac{2}{3}-\dfrac{1}{24}[/tex]
[tex]p=\dfrac{-16-1}{24}[/tex]
[tex]p=\dfrac{-17}{24}[/tex]
Putting [tex]h=-\dfrac{1}{4},k=\dfrac{1}{24}, p=-\dfrac{17}{24}[/tex] in (i), we get
[tex]\left(x-(-\dfrac{1}{4})\right)^2=4\left(-\dfrac{17}{24}\right)\left(y-\dfrac{1}{24}\right)[/tex]
[tex]\left(x+\dfrac{1}{4}\right)^2=-\dfrac{17}{6}\left(y-\dfrac{1}{24}\right)[/tex]
[tex]x^2+2(x)(\dfrac{1}{4})+(\dfrac{1}{4})^2=-\dfrac{17}{6}\left(y-\dfrac{1}{24}\right)[/tex]
[tex]-\dfrac{6}{17}\left(x^2+\dfrac{1}{2}x+\dfrac{1}{16}\right)=y-\dfrac{1}{24}[/tex]
On further simplification, we get
[tex]-\dfrac{6}{17}(x^2)-\dfrac{6}{17}(\dfrac{1}{2}x)-\dfrac{6}{17}(\dfrac{1}{16})=y-\dfrac{1}{24}[/tex]
[tex]-\dfrac{6}{17}x^2-\dfrac{3}{17}x-\dfrac{3}{136}+\dfrac{1}{24}=y[/tex]
[tex]-\dfrac{6}{17}x^2-\dfrac{3}{17}x+\dfrac{-9+17}{408}=y[/tex]
[tex]-\dfrac{6}{17}x^2-\dfrac{3}{17}x+\dfrac{8}{408}=y[/tex]
[tex]-\dfrac{6}{17}x^2-\dfrac{3}{17}x+\dfrac{1}{51}=y[/tex]
Therefore, the equation of the parabola is [tex]y=-\dfrac{6}{17}x^2-\dfrac{3}{17}x+\dfrac{1}{51}[/tex]. Hence, the correct option is C.
