A N S W E R Q U I C K P L E A S E 1. The directrix of a parabola is y=−8 . The focus of the parabola is (−2,−6) .What is the equation of the parabola?y=14(x+2)2−7y=−18(x+2)2+7y=18(x−2)2−7y=−14(x−2)2−72. The directrix of a parabola is the line y=5 . The focus of the parabola is (2,1) .What is the equation of the parabola?y=−18(x−2)2−3y=18(x−2)2+3y=18(x−2)2−3y=−18(x−2)2+33. The focus of a parabola is (0,−2) . The directrix of the parabola is the line y=−3 .What is the equation of the parabola?y=14x2−2y=−12x2−52y=−14x2+2y=12x2−52
Accepted Solution
A:
Answer:1. A2. D3. DStep-by-step explanation:The standard form of a parabola is[tex]y=\frac{1}{4p}(x-h)^2+k[/tex] ..... (1)Where, (h,k) is vertex, (h,k+p) is focus and y=k-p is directrix.1. The directrix of a parabola is y=−8 . The focus of the parabola is (−2,−6) .[tex]k-p=-8[/tex] ...(a)[tex](h,k+p)=(-2,-6)[/tex][tex]k+p=-6[/tex] .... (b)[tex]h=-2[/tex]On solving (a) and (b), we get k=-7 and p=1.Put h=-2, k=-7 and p=1 in equation (1).[tex]y=\frac{1}{4(1)}(x-(-2))^2+(-7)[/tex][tex]y=\frac{1}{4}(x+2)^2-7[/tex]Therefore option A is correct.2 The directrix of a parabola is the line y=5 . The focus of the parabola is (2,1) .[tex]k-p=5[/tex] ...(c)[tex](h,k+p)=(2,1)[/tex][tex]k+p=1[/tex] .... (d)[tex]h=2[/tex]On solving (c) and (d), we get k=3 and p=-2.Put h=2, k=3 and p=-2 in equation (1).[tex]y=\frac{1}{4(-2)}(x-(2))^2+(3)[/tex][tex]y=-\frac{1}{8}(x-2)^2+3[/tex]Therefore option D is correct.3. The focus of a parabola is (0,−2) . The directrix of the parabola is the line y=−3 .[tex]k-p=-3[/tex] ...(e)[tex](h,k+p)=(0,-2)[/tex][tex]k+p=-2[/tex] .... (f)[tex]h=0[/tex]On solving (e) and (f), we get k=-2.5 and p=0.5.Put h=0, k=-2.5 and p=0.5 in equation (1).[tex]y=\frac{1}{4(0.5)}(x-(0))^2+(-2.5)[/tex][tex]y=\frac{1}{2}(x)^2-2.5[/tex][tex]y=\frac{1}{2}(x)^2-\frac{5}{2}[/tex]Therefore option D is correct.