The long axis of the ellipse is known to be 4 in length and passes through the point root 3, 1 2 .

a=2 If the focus is on the x-axis, set the elliptic equation: x 4 + y b =1 through the point (root 3, 1 2).

3 4 + 1 (4b )=1 The elliptic equation for b =1 is: x 4 + y =1

If the focus is on the y-axis, set the elliptic equation: y4 + x b =1 through the point (root 3, 1 2).

1 16 + 3 b =1 b =16 5 The elliptic equation is: y 4 + 5x 16 = 1

Solution: (1) Let the equation of the ellipse be x2 a2 + y2 b2 = 1, because the length of the major axis is 4, then a = 2, get a2 = 4, and because the ellipse passes through the point ( 3, 1 2), so bring x = 3, y = 1 2 and a2 = 4 into x2 a2 + y2 b2 = 1, get b2 = 1, so the elliptic equation is x2 4 + y2 = 1

2) Let the equation of the ellipse be x2 b2 + y2 a2 = 1, because the length of the major axis is 4, that is, the absolute value of 2a is 4, then a = 2, get a2 = 4, and because the ellipse passes through the point ( 3, 1 2), so bring x = 3, y = 1 2 and a2 = 4 into x2 b2 + y2 a2 = 1, get b2 = 16 5, so the elliptic equation is 5x2 16 + y2 4 = 1

In summary, the equation for the chosen ellipse is x2 4 + y2 = 1 or 5x2 16 + y2 4 = 1.

It can be known that a is 2, list the elliptic equation, bring 2 in, and then bring in the coordinates of the passing point.

The length of the major axis is 4, the root number 2 2a=4 2 a=2 2 x 2 a 2+y 2 b 2=1 and the point (-root number 2,2) (2) 2 bend (2 2) 2+4 b 2=1 b 2=16 3 focus on the x-axis elliptic standard equation x 2 8+3y 2 lack of lap 16=1b=2 22 a 2+4 8=1a 2=4 focus on the y-axis elliptic standard equation volt-slip x 2 4+y 2 8=1....

Ellipse: x 2 4 + y 2 9 = 1, the major axis is on the y axis, the length is: 6, according to the problem the standard equation for the ellipse is:

x 2 a 2 + y 2 9 = 1, substitution point (-4, 1) to obtain: (-4) 2 a 2 + 1 2 9 = 1, (a>3).

The solution is: a 2 = 18, so the standard equation for the ellipse is x 2 18 + y 2 9 = 1.

The standard equation for finding an ellipse is that the length of the minor axis is 6 and the crossing point (1,4).

Solution: (a). When the elliptic focus is on the x-axis, since b=3, the elliptic equation can be set to x a +y 9=1, and the points (1,4) can be substituted.

1 a +16 9 = 1, 1 a = 1-16 9 = -7 9, obviously there is no solution, that is, the focus cannot be on the x-axis.

ii). When the elliptic focus is on the y-axis, the elliptic equation can be set to y a +x 9=1, and the points (1,4) are substituted to obtain 16 a +1 9=1

16 a = 1-1 9 = 8 9, so a = 16 (9 8) = 18, so the elliptic equation is y 18 + x 9 = 1

Let the equation be x 2 m+y 2 n=1

If m=(6 2) 2=9 then 1 9+y 2 n=1 => n=18

If n=9 then 1 m+16 9=1 => m<0 [not in accordance with the topic, discarded.] 】

The standard equation for an ellipse is x 2 9+y 2 18=1

So be cautious a=2b

Let x 4b +y b = 1

The coordinates of the base are obtained as b=2 3, a=4 3

Then Pipe gets the equation: 9x 16+9y 4=1

x a + y b = 1 standard equation.

The major axis of an ellipse is 3 times larger than the minor axis.

a=3b or b=3a

Substitution x 9b +y b =1 or x a +y 9a =1 through the dot p(3,0).

x Shenwu 9b +y b =1 or x a +y 9a = 1

x 9+y =1 or x 9+y clump thick 81=1

The length of the balance axis of the short jujube is 2 5

2b=2√5

b = 5 when the elliptic equation is x a + y 5 = 1 is substituted by ( 3,-2).

a = 15a = 15>b is true.

The elliptic equation is x 15 + y 5 = 1

When the elliptic equation is y a +x 5=1, substitute the prudence (3,-2) into it.

a = 10a = 10>b also holds.

The elliptic equation is y 10 + x 5 = 1

Solution: Because 2a=4, a=2; 2b=1 2*2a, so b=1;

Discussion of Divination: If the focus is on the x-axis, then the equation x 2 4+y 2 1=1;

If the focus is on the y-axis, then the equation is this roll: y2 4+x 2, 1=1