The equation of the ellipse with center \(\displaystyle (h,k)\), horizontal axis of length \(\displaystyle 2a\), and vertical axis of length \(\displaystyle 2b\) is

\(\displaystyle \frac{(x-h)^{2}}{a^{2}} + \frac{(y-k)^{2}}{b^{2}} = 1\)

The ellipse has center \(\displaystyle (6,2)\), horizontal axis of length 8, and vertical axis of length 16. Therefore,

\(\displaystyle h = 6,k = 2\), \(\displaystyle a = \frac{8}{2} = 4\), and \(\displaystyle b = \frac{16}{2} = 8\).

The equation of the ellipse is

\(\displaystyle \frac{(x-6)^{2}}{4^{2}} + \frac{(y-2)^{2}}{8^{2}} = 1\)

\(\displaystyle \frac{(x-6)^{2}}{16} + \frac{(y-2)^{2}}{64} = 1\)

The equation of the ellipse with center \(\displaystyle (h,k)\), horizontal axis of length \(\displaystyle 2a\), and vertical axis of length \(\displaystyle 2b\) is

\(\displaystyle \frac{(x-h)^{2}}{a^{2}} + \frac{(y-k)^{2}}{b^{2}} = 1\)

The ellipse has center \(\displaystyle (1,1)\), horizontal axis of length 10, and vertical axis of length 6. Therefore,

\(\displaystyle h = k = 1\), \(\displaystyle a = \frac{10}{2} = 5\), and \(\displaystyle b = \frac{6}{2} = 3\).

The equation of the ellipse is

\(\displaystyle \frac{(x-1)^{2}}{5^{2}} + \frac{(y-1)^{2}}{3^{2}} = 1\)

\(\displaystyle \frac{(x-1)^{2}}{25} + \frac{(y-1)^{2}}{9} = 1\)

The equation of the ellipse with center \(\displaystyle (h,k)\), horizontal axis of length \(\displaystyle 2a\), and vertical axis of length \(\displaystyle 2b\) is

\(\displaystyle \frac{(x-h)^{2}}{a^{2}} + \frac{(y-k)^{2}}{b^{2}} = 1\)

The ellipse has center \(\displaystyle (3,5)\), horizontal axis of length 8, and vertical axis of length 6. Therefore,

\(\displaystyle h = 3,k = 5\), \(\displaystyle a = \frac{8}{2} = 4\), and \(\displaystyle b = \frac{6}{2} = 3\).

The equation of the ellipse is

\(\displaystyle \frac{(x-3)^{2}}{4^{2}} + \frac{(y-5)^{2}}{3^{2}} = 1\)

\(\displaystyle \frac{(x-3)^{2}}{16} + \frac{(y-5)^{2}}{9} = 1\)

To find the \(\displaystyle y\)-intercept, that is, the point of intersection with the \(\displaystyle y\)-axis, of the line of equation \(\displaystyle 3x+ 5y = 30\), set \(\displaystyle x=0\) and solve for \(\displaystyle y\):

\(\displaystyle 3 (0)+ 5y = 30\)

\(\displaystyle 5y = 30\)

\(\displaystyle y = 6\)

The \(\displaystyle y\)-intercept is \(\displaystyle (0,6)\).

The \(\displaystyle x\)-intercept can be found by doing the opposite:

\(\displaystyle 3x+ 5 (0) = 30\)

\(\displaystyle 3x = 30\)

\(\displaystyle x=10\)

The \(\displaystyle x\)-intercept is \(\displaystyle (10,0)\).

The parabola has these intercepts as well. Also, since the vertical parabola has only one \(\displaystyle x\)-intercept, that point doubles as its vertex as well. 

The equation of a vertical parabola, in vertex form, is

\(\displaystyle y = a(x-h)^{2}+k\),

where \(\displaystyle (h,k)\) is the vertex. Set \(\displaystyle h = 10,k=0\):

\(\displaystyle y = a(x-10)^{2}+0\)

\(\displaystyle y = a(x-10)^{2}\)

for some real \(\displaystyle a\). To find it, use the \(\displaystyle y\)-intercept, setting \(\displaystyle x=0, y = 6\)

\(\displaystyle 6 = a(0-10)^{2}\)

\(\displaystyle 6 = a( -10)^{2}\)

\(\displaystyle 6 = 100a\)

\(\displaystyle a = \frac{6}{100} = \frac{3}{50}\)

The parabola has equation \(\displaystyle y =\frac{3}{50} (x-10)^{2}\), which is rewritten as

\(\displaystyle y =\frac{3}{50} (x^{2}-20x+100)\)

\(\displaystyle y = \frac{3}{50} x^{2}-\frac{6}{5}x+6\)

The equation of the ellipse with center \(\displaystyle (h,k)\), horizontal axis of length \(\displaystyle 2a\), and vertical axis of length \(\displaystyle 2b\) is

\(\displaystyle \frac{(x-h)^{2}}{a^{2}} + \frac{(y-k)^{2}}{b^{2}} = 1\)

The center is \(\displaystyle (-4, 9)\), so \(\displaystyle h = -4\) and \(\displaystyle k = 9\).

To find \(\displaystyle a\), note that one endpoint of the horizontal axis is given by the point with the same \(\displaystyle y\)-coordinate through which it passes, namely, \(\displaystyle (11, 9)\). Half the length of this axis, which is \(\displaystyle a\), is the difference of the \(\displaystyle x\)-coordinates, so \(\displaystyle a = 11- (-4) = 15\). Similarly, to find \(\displaystyle b\), note that one endpoint of the vertical axis is given by the point with the same \(\displaystyle x\)-coordinate through which it passes, namely, \(\displaystyle (-4, 5)\). Half the length of this axis, which is \(\displaystyle b\), is the difference of the \(\displaystyle x\)-coordinates, so \(\displaystyle b = 9 - 5 = 4\).

The equation is 

\(\displaystyle \frac{(x-(-4))^{2}}{15^{2}} + \frac{(y-9)^{2}}{4^{2}} = 1\)

or 

\(\displaystyle \frac{(x+4)^{2}}{225} + \frac{(y-9)^{2}}{16} = 1\).

The equation of the ellipse with center \(\displaystyle (h,k)\), horizontal axis of length \(\displaystyle 2a\), and vertical axis of length \(\displaystyle 2b\) is

\(\displaystyle \frac{(x-h)^{2}}{a^{2}} + \frac{(y-k)^{2}}{b^{2}} = 1\)

\(\displaystyle (4, 6)\) and \(\displaystyle (10,6)\) are the endpoints of a horizontal line segment with midpoint

\(\displaystyle \left ( \frac{4+10}{2}, \frac{6+6}{2} \right )\), or \(\displaystyle (7,6)\)

and length \(\displaystyle 10-4= 6\).

\(\displaystyle (7, 2)\) and \(\displaystyle (7, 10)\) are the endpoints of a vertical line segment with midpoint

\(\displaystyle \left ( \frac{7+7}{2}, \frac{2+10}{2} \right )\), or \(\displaystyle (7,6)\)

and length \(\displaystyle 10-2 = 8\)

Because their midpoints coincide, these are the endpoints of the horizontal axis and vertical axis, respectively, of the ellipse, and the common midpoint \(\displaystyle (7,6)\) is the center.

Therefore, 

\(\displaystyle h = 7\) and \(\displaystyle k = 6\);

\(\displaystyle 2a = 6\) and \(\displaystyle 2b= 8\); consequently \(\displaystyle a = 3\) and \(\displaystyle b = 4\).

The equation of the ellipse is 

\(\displaystyle \frac{(x-7)^{2}}{3^{2}} + \frac{(y-6)^{2}}{4^{2}} = 1\), or

\(\displaystyle \frac{(x-7)^{2}}{9} + \frac{(y-6)^{2}}{16} = 1\)

If a horizontal parabola has only one \(\displaystyle y\)-intercept, which here is \(\displaystyle (0,4)\), that point doubles as its vertex as well. 

The equation of a horizontal parabola, in vertex form, is

\(\displaystyle x = a(y-k)^{2}+h\),

where \(\displaystyle (h,k)\) is the vertex. Set \(\displaystyle h = 0,k=4\):

\(\displaystyle x = a(y-4)^{2}+0\)

\(\displaystyle x = a(y-4)^{2}\)

To find \(\displaystyle a\), use the \(\displaystyle x\)-intercept, setting \(\displaystyle x= 6, y=0\):

\(\displaystyle 6 = a(0-4)^{2}\)

\(\displaystyle 6 = a( -4)^{2}\)

\(\displaystyle 6 = 16a\)

\(\displaystyle a = \frac{6}{16} = \frac{3}{8}\)

The equation, in vertex form, is \(\displaystyle x = \frac{3}{8}(y-4)^{2}\). In standard form:

\(\displaystyle x = \frac{3}{8}(y^{2}-8y+16)\)

\(\displaystyle x = \frac{3}{8} y^{2}-3y+6\)

The equation of a horizontal parabola, in vertex form, is

\(\displaystyle x = a(y-k)^{2}+h\),

where \(\displaystyle (h,k)\) is the vertex. Set \(\displaystyle h = -3,k=-6\):

\(\displaystyle x = a[y-(-6)]^{2}+ (-3)\)

\(\displaystyle x = a(y+6)^{2}-3\)

To find \(\displaystyle a\), use the known \(\displaystyle y\)-intercept, setting \(\displaystyle x =0, y=4\):

\(\displaystyle 0 = a(4+6)^{2}-3\)

\(\displaystyle 0 = a\cdot 10^{2}-3\)

\(\displaystyle 0 =100 a-3\)

\(\displaystyle 100a= 3\)

\(\displaystyle a = \frac{3}{100}\)

The equation, in vertex form, is \(\displaystyle x = \frac{3}{100}(y+6)^{2}-3\); in standard form:

\(\displaystyle x = \frac{3}{100}(y^{2}+12+36)-3\)

\(\displaystyle x = \frac{3}{100} y^{2}+\frac{9}{25} y+\frac{27}{25} - \frac{75}{25}\)

\(\displaystyle x = \frac{3}{100} y^{2}+\frac{9}{25} y-\frac{48}{25}\)

A horizontal parabola which passes through  \(\displaystyle (0,2)\) and \(\displaystyle (0,6)\) has as its equation

\(\displaystyle x = a(y-2)(y-6)\).

To find \(\displaystyle a\), substitute the coordinates of the third point, setting \(\displaystyle x=3,y=-2\):

\(\displaystyle 3= a(-2-2)(-2-6)\)

\(\displaystyle 3= a(-4)(-8)\)

\(\displaystyle 3= 32a\)

\(\displaystyle a = \frac{3}{32}\)

The equation is therefore \(\displaystyle x = \frac{3}{32}(y-2)(y-6)\), which is, in standard form:

\(\displaystyle x = \frac{3}{32}(y^{2}-8y+12)\)

\(\displaystyle x = \frac{3}{32} y^{2}- \frac{3}{4} y+ \frac{9}{8}\)

A vertical parabola which passes through \(\displaystyle (2,0)\) and \(\displaystyle (6,0)\) has as its equation

\(\displaystyle y = a(x-2)(x-6)\)

To find \(\displaystyle a\), substitute the coordinates of the third point, setting \(\displaystyle x=3,y=-2\):

\(\displaystyle -2 = a(3-2)(3-6)\)

\(\displaystyle -2 = a \cdot 1 \cdot (-3)\)

\(\displaystyle -2 = -3a\)

\(\displaystyle a = \frac{-2}{-3}= \frac{2}{3}\)

The equation is \(\displaystyle y = \frac{2}{3}(x-2)(x-6)\); expand to put it in standard form:

\(\displaystyle y = \frac{2}{3}(x^{2}-8x+12)\)

\(\displaystyle y = \frac{2}{3}x^{2}-\frac{16}{3}x+8\)