Question

Select the correct equations in the image.

Based on the the equations of the parabolas, identify each parabola whose focus and vertex lie in different quadrants.

Answer 1

Answer:

The equations of the parabolas whose focus and vertex lie in different quadrants are;

The third equation $x=-\frac{y^{2}}{16}-\frac{y}{4}+\frac{11}{4}$ and the sixth equation$y=-\frac{x^{2}}{24}-\frac{5x}{12}+\frac{95}{24}$

Step-by-step explanation:

We analyse each of the equations as follows

$h =\frac{-b}{2a}$

$k = \frac{4ac-b^2}{4a}$

Vertex = V(h, k)

Focus = F(h, k+p)

First equation

a = -1/8

b = 1/4

c = 23/8

k = 3

h = 1 Hence same quadrant

Second equation

a = 1/32

b = 1/4

c = -13/2

h = -4

k = -7 same quadrant

Third equation

a = -1/16

b = -1/4

c = 11/4

k = 3

h = -2 Hence different quadrants

Fourth equation

a = 1/16

b = 1/4

c = -19/4

h = -2

k = -5 The same quadrant

Fifth equation

a = -1/36

b = -5/18

c = 299/36

k = 620.56

h = 149.5 The same quadrant and

Sixth equation

a = -1/24

b = -5/12

c = 95/24

h = -5

k ≈ 5   Hence different quadrants

Hence the equations of the parabolas whose focus and vertex lie in different quadrants are the third and sixth equations presented as follows;

Third equation:

$x=-\frac{y^{2}}{16}-\frac{y}{4}+\frac{11}{4}$

Sixth equation:

$y=-\frac{x^{2}}{24}-\frac{5x}{12}+\frac{95}{24}$.

Answer 2

The equations of the parabolas, whose focus and vertex lie in different quadrants, are,

$x=-\dfrac{y^2}{16}-\dfrac{y}{4}+\dfrac{11}{4}$

$x=-\dfrac{y^2}{36}-\dfrac{5y}{18}+\dfrac{299}{36}$

What is equations of the parabolas?

The equation of parabola is the way to represent a parabola in a algebraic expression from using its vertex points.

The general form of the equation of the parabola can be given as,

$y=a(x-h)^2+k$

Here, (h, k) are the vertex. These can be given as,

$h=\dfrac{-b}{2a}\\k=\dfrac{4ac-b^2}{4a}$

Let's check all the options, whose vertex lie in the different quadrant, using the above formula,

The first equation given in the problem is,

$y=-\dfrac{x^2}{8}+\dfrac{x}{4}+\dfrac{23}{8}$

Here, a is -1/8, b is 1/4 and c is 23/8. By these points, the vertex point we get as,

$h=\dfrac{-\dfrac{1}{4}}{(2)\dfrac{-1}{8}}=3\\k=\dfrac{4(-1/8)(23/8)-(1/4)^2}{4(-1/8)}=1$

Both vertex has positive sign, (3,1). Thus they are in the same quadrant.

Similarly, for the second equation,

$y=-\dfrac{x^2}{32}+\dfrac{x}{4}-\dfrac{13}{2}$

a is 1/32, b is 1/4 and c is -13/2. By these points, the vertex point we get as,

$h=-4\\k=-7$

Both vertex has positive sign, (-4,-7). Thus, they are in the same quadrant.

For the third equation,

$x=-\dfrac{y^2}{16}-\dfrac{y}{4}+\dfrac{11}{4}$

Here, a is -1/16, b is -1/4 and c is 11/4. By these points, the vertex point we get as,

$h=3\\k=-2$

Both vertex has opposite sign, (3,-2). Thus, they are in the different quadrant.

For the fourth equation,

$x=\dfrac{y^2}{16}+\dfrac{y}{4}-\dfrac{19}{4}$

a is 1/16, b is 1/4 and c is -19/4. By these points, the vertex point we get as,

$h=-2\\k=-5$

Both vertex has positive sign, (-2,-5). Thus, they are in the same quadrant.

For the fifth equation,

$x=-\dfrac{y^2}{36}-\dfrac{5y}{18}+\dfrac{299}{36}$

Here, a is -1/36, b is -5/18 and c is 299/36. By these points, the vertex point we get as,

$h=620.56\\k=\dfrac{299}{2}$

Both vertex has positive sign, (620.56, 299/2). Thus, they are in the same quadrant.

For the sixth equation, a is -1/24, b is -5/12 and c is 95/24. By these points, the vertex point we get as,

$h=-5\\k=5$

Both vertex has opposite sign, (-5,5).  Thus, they are in the different quadrant.

Hence, the equations of the parabolas, whose focus and vertex lie in different quadrants, are,

$x=-\dfrac{y^2}{16}-\dfrac{y}{4}+\dfrac{11}{4}$

$x=-\dfrac{y^2}{36}-\dfrac{5y}{18}+\dfrac{299}{36}$

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