Question #157470

Write down the standard form of the equation of a parabola in a given directrix and vertex at (0, 0)

(a) x = -4

(b) y = 2

(c) y = - 3

Expert's answer

We can write the equation of parabola in the form:

4p(y−k)=(x−h)24p(y-k)=(x-h)^2 . The focus is (h,k+p) and the directrix is y=k−py=k-p. The vertex is (h,k). In item b) we receive: k−p=2k-p=2 and h=0,k=0h=0, k=0. Thus, p=−2p=-2. The equation is: −8y=x2-8y=x^2

In c). we get k−p=−3k-p=-3, h=0,k=0h=0,k=0. The equation is: 12y=x212y=x^2

Now we write the equation in the form: 4p(x−k)=(y−h)24p(x-k)=(y-h)^2 . The focus is (h,k+p) and the directrix is x=k−px=k-p. The vertex is (h,k). For a). we receive: k=0,h=0k=0,h=0, k−p=−4k-p=-4. Thus, p=4.p=4. The equation is: 16x=y216x=y^2 . The standard forms of equations of parabola are: y=ax2+bx+cy=ax^2+bx+c and x=ey2+fy+gx=ey^2+fy+g , where a,b,c,e,f,g∈Ra,b,c,e,f,g\in{\mathbb{R}}. All obtained equations are in standard forms.

Answer: a). 16x=y216x=y^2; b). −8y=x2-8y=x^2 ; c). 12y=x212y=x^2



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