25 Points Question 21 of 40 Convert each equation to standar

2.5 Points Question 21 of 40 Convert each equation to standard form by completing the the parabola. square on x or y. Then find the vertex, focus, and directrix of OA, (x-4)2 = 4(y-2): vertex: (1,4); focus: (1.3) ; directrix: y-1 OB, (x-2)2 4(y-3); vertex: (1,2); focus: (1, 3) ; directrix: y = 3 Oc. (x-1F-4(y . 2); vertex: (1,2); focus: (1, 3) ; directrix: y-1 O D.(x-p?#2(y-2): vertex: (1,3); focus: (1, 2) ; directrix: y-5

Solution

(21)

x² - 2x - 4y + 8 = 0 is the given equation of the parabola.

Now by completing the square method it can be re-written as -

(x - 1)² - 4y + 8 = 0

or (x - 1)² = 4(y - 2)         ......(1)

Now compare this equation with the standard form of a parabola (x - h)² = 4a(y - k). We know that the vertex is where the x and y terms are. we find the x-coordinate is 1 and the y-coordinate is 2.

So Vertex (V) = (1,2)

Now to find focus and directrix of the parabola we need to compare 4a with the term in front of the y term (in parenthesis) so 4a = 4 which means a = 1. it is the distance of focus from the vertex. Now since this parabola is opening in the +ve direction of the y-axis and focus is within the curve, so focus is (1,3). The directrix is equidistant from the vertex that the focus is but outside the curve in down direction, so directrix is y=1.

So option (C) is your answer.

(22)

Given parabola is (y + 3)² = 12(x + 1)

To find Vertex, Focus and Directrix, we have to follow the same process as explained in answer (21) above. It\'s a parabola opening in the +ve direction of the x-axis.

Vertex = (-1,-3), focus = (2,-3) and directrix is x = -4

So option (D) is your answer.

(30)

The given equation of the ellipse is > 7x² + 5y² = 35 can be rewritten as -

divide the whole equation by 35, we get x²/5 + y²/7 = 1

Since the denominator of y²/7 is larger than the denominator of x²/5 so the major axis is along the y-axis. now pon comparing this equation of ellipse with standard equation x²/b² + y²/a² = 1, we have b² = 5 and a² = 7.

Now from the formula c = (a² – b²) = (7 - 5) = 2

Hence the foci are (0, 2) and (0, -2)

So option (A) is your answer.

(31)

The given equation of the hyperbola is y²/4 - x²/1 = 1

Since the equation of the parabola is of the form y²/a² - x²/b² = 1 which has it vertices and foci on y-axis.

Upon comparing we have b² = 1 and a² = 4

Now from the formula c = (a² + b²) = (4 + 1) = 5

Hence the foci are (0, 5) and (0, -5)
and Vertices are (0,2) and (0, -2) by a² = 4 giving a = ±2

So option (C) is your answer.