Please answer both parts A & B thanks
Determine the equation of each of the following a parabola with focus (2, 1) and directrix y = -2 an ellipse with foci (0, 2) and (0. -2) with y Intercepts at (0, 3) and (0, -3)
Solution
A. This is a parabola with a vertical axis of symmetry, since the directrix is a horizontal line y = -2 (h,k) are the coordinates of the focus. Let P(x,y) be any point on the parabola, then P is equidistant from the focus and the directrix. Apply Pythagoras for the distance between the focus and the curve. Since this distance is equal to the distance from the directrix to the curve then:- sqrt[(x - 2)^2 + (y - 1)^2] = y + 2 Pythagorean part. Directrix part. Square up [(x - 2)^2 + (y - 1)^2] = (y + 2)^2 Expand x^2 - 4x + 4 + y^2 - 2y +1 = y^2 + 4y + 4 Collect like terms and equate to \'0\' x^2 - 4x + 4 + y^2 - 2y + 1 - y^2 - 4y - 4 = 0 x^2 - 4x + 4 - 6y - 3 = 0 6y = x^2 - 4x + 1 or y = x^2/6 - 2x/6 + 1/6 or y = x^2/6 - x/3 + 1/6 B.Standard form of ellipse with horizontal major axis: (x-h)^2/a^2+(y-k)^2/b^2=1, (a>b), with (h,k) being the (x,y) coordinates of the center. Standard form of ellipse with vertical major axis: (x-h)^2/b^2+(y-k)^2/a^2=1, (a>b), with (h,k) being the (x,y) coordinates of the center. The difference between the two forms is the interchange of a^2 and b^2 .. Because the given x-coordinates of foci and vertices are the same at x=0, ellipse has a vertical major axis. length of major axis=8=2a center at (0,0) (midpoints of vertices and foci) a=4 a^2=16 c=half the distance between foci=4/2=2 c^2=4 c^2=a^2-b^2 b^2=a^2-c^2=16-4=12 b=v12=3.46 .. Equation:(x-0)^2/12+(y-0)^2/16=1 =x^2/12+y^2/16=1
.
HMCS Max Bernays Pre-Deployment Brief (May 2024).pptx
Please answer both parts A & B thanks Determine the equation of each o.docx
1. Please answer both parts A & B thanks
Determine the equation of each of the following a parabola with focus (2, 1) and directrix y = -2
an ellipse with foci (0, 2) and (0. -2) with y Intercepts at (0, 3) and (0, -3)
Solution
A. This is a parabola with a vertical axis of symmetry, since the directrix is a horizontal line y = -
2 (h,k) are the coordinates of the focus. Let P(x,y) be any point on the parabola, then P is
equidistant from the focus and the directrix. Apply Pythagoras for the distance between the focus
and the curve. Since this distance is equal to the distance from the directrix to the curve then:-
sqrt[(x - 2)^2 + (y - 1)^2] = y + 2 Pythagorean part. Directrix part. Square up [(x - 2)^2 + (y -
1)^2] = (y + 2)^2 Expand x^2 - 4x + 4 + y^2 - 2y +1 = y^2 + 4y + 4 Collect like terms and
equate to '0' x^2 - 4x + 4 + y^2 - 2y + 1 - y^2 - 4y - 4 = 0 x^2 - 4x + 4 - 6y - 3 = 0 6y = x^2 - 4x
+ 1 or y = x^2/6 - 2x/6 + 1/6 or y = x^2/6 - x/3 + 1/6 B.Standard form of ellipse with horizontal
major axis: (x-h)^2/a^2+(y-k)^2/b^2=1, (a>b), with (h,k) being the (x,y) coordinates of the
center. Standard form of ellipse with vertical major axis: (x-h)^2/b^2+(y-k)^2/a^2=1, (a>b), with
(h,k) being the (x,y) coordinates of the center. The difference between the two forms is the
interchange of a^2 and b^2 .. Because the given x-coordinates of foci and vertices are the same at
x=0, ellipse has a vertical major axis. length of major axis=8=2a center at (0,0) (midpoints of
vertices and foci) a=4 a^2=16 c=half the distance between foci=4/2=2 c^2=4 c^2=a^2-b^2
b^2=a^2-c^2=16-4=12 b=v12=3.46 .. Equation:(x-0)^2/12+(y-0)^2/16=1 =x^2/12+y^2/16=1
![Please answer both parts A & B thanks
Determine the equation of each of the following a parabola with focus (2, 1) and directrix y = -2
an ellipse with foci (0, 2) and (0. -2) with y Intercepts at (0, 3) and (0, -3)
Solution
A. This is a parabola with a vertical axis of symmetry, since the directrix is a horizontal line y = -
2 (h,k) are the coordinates of the focus. Let P(x,y) be any point on the parabola, then P is
equidistant from the focus and the directrix. Apply Pythagoras for the distance between the focus
and the curve. Since this distance is equal to the distance from the directrix to the curve then:-
sqrt[(x - 2)^2 + (y - 1)^2] = y + 2 Pythagorean part. Directrix part. Square up [(x - 2)^2 + (y -
1)^2] = (y + 2)^2 Expand x^2 - 4x + 4 + y^2 - 2y +1 = y^2 + 4y + 4 Collect like terms and
equate to '0' x^2 - 4x + 4 + y^2 - 2y + 1 - y^2 - 4y - 4 = 0 x^2 - 4x + 4 - 6y - 3 = 0 6y = x^2 - 4x
+ 1 or y = x^2/6 - 2x/6 + 1/6 or y = x^2/6 - x/3 + 1/6 B.Standard form of ellipse with horizontal
major axis: (x-h)^2/a^2+(y-k)^2/b^2=1, (a>b), with (h,k) being the (x,y) coordinates of the
center. Standard form of ellipse with vertical major axis: (x-h)^2/b^2+(y-k)^2/a^2=1, (a>b), with
(h,k) being the (x,y) coordinates of the center. The difference between the two forms is the
interchange of a^2 and b^2 .. Because the given x-coordinates of foci and vertices are the same at
x=0, ellipse has a vertical major axis. length of major axis=8=2a center at (0,0) (midpoints of
vertices and foci) a=4 a^2=16 c=half the distance between foci=4/2=2 c^2=4 c^2=a^2-b^2
b^2=a^2-c^2=16-4=12 b=v12=3.46 .. Equation:(x-0)^2/12+(y-0)^2/16=1 =x^2/12+y^2/16=1](https://image.slidesharecdn.com/pleaseanswerbothpartsabthanksdeterminetheequationofeacho-230131011507-f96a3353/85/Please-answer-both-parts-A-B-thanks-Determine-the-equation-of-each-o-docx-1-320.jpg)
