Question

halla la ecuacion ordinaria de una parabola dados V(4,3); d: x=0

Answer 1

$(x-h)^2 = 4p(y-k)\qquad \\ \\ (x-4)^2 = 4p(y-3)\qquad p= (4-3) = 1 \\ \\ (x-4)^2 = 4(1)(y-3) \\ \\ x^2-8x+16= 4y-12 \\ \\ x^2-8x+16+12= 4y \\ \\ x^2-8x+28= 4y \\ \\ \frac{x^2}{4} - \frac{8x}{4}+ \frac{28}{4}= y \\ \\ \boxed{y= \frac{1}{4} x^2-2x +7}$


Espero que te sirva, salu2!!!!