Math If the equation y2 x 2lga 1 3 a represents an ellipse with the focus on the x axis

If the equation y2-x 2LGA=1 3-a represents the ellipse with the focal point on the x-axis, then the range of the real number a is

Just be satisfied. lga 0, and -lga 1

and 1 3-a 0, that's it;

Solution: 1 109 meets the requirements of the question.

When elliptical, xx (25-k)+yy (9-k)=1 becomes xx (25-k)+yy(9-k)=1, then there is 25-k-9+k=16=25-9, i.e., k<9

If it is hyperbola, it is necessary to ensure that the coefficients of xx 2 and y 2 are different.

, you need to meet: |k|-2>0 ,5-k<0

or |k|-2<0,5-k>0

1-k^2)x^2 + 2(2k-3)x -(4k^2 -12k +13) =0;

Since there are two different real solutions, then the above equation discriminant = [2(2k-3)] 2 -4(1-k 2)· (4k 2 -12k +13) 0;

And from the range of the equation x -2 or x 2, it is concluded that

The value range of the real number k is (-1,3 4].

The focus is on the x-axis, then -1 LGA>1 then we get >1a> and the right side of the equal sign should be greater than 0

So less than 1 3

After simplification, it is (x-1) 2 9+(y-2) 2 This is an ellipse that is not in the standard position, and the center is (1,2) so just go straight through (1,2) and you will do the rest.

According to the problem, it is easy to find 25-k-(9-k)=16, so as long as it is an elliptic equation, then 25-k>0 9-k>0 can also find 25-k+(k-9)=16

So... Know?

Since it is hyperbola, then (5-k)*+k|-2)<0 (5-k)(|k|-2) is not equal to 0

Just do the math.

Didn't come up with it.

x2 comic (丨m丨-3) +y2 respectful finger segment (5-m)=1 indicates the ellipse focused on the y-axis.

Then 5-m>imi-3

imi+m

The tangent of the elliptic imaginary lead at x-y+5=0 is, and we find b

3x^2+4y^2=12m

In y=x+5 column of equations, and the discriminant formula (triangle) = 0 find m and write the standard equation on which basis can be Li Yujin.

Because it is an elliptic equation, k>0

into a standard equation.

x^2/2+y^2/(2/k)=1

The focus is on the y-axis.

2/k>2

The value range of the real number k is (0,1) for 1 k>1 0

Solution: Equation x (25-m) + y (m+9) = 1There should be m+9 25-m 0===>8＜m＜25.

x 2 + my 2 = 1 can be converted to: x 2 + y 2 1 m = 1 because the figure is an ellipse, so 1 m cannot be equal to 1, if 1 m is equal to 1, then the above formula = x 2 + y 2 = 1 is the round rock know the beat!

If the point is on the y-axis, then 1 m<1, which means that the fierce bridge is m<1

Again, in x 2 a 2 + y 2 b 2 = 1, the value cannot be equal to 0Therefore, m cannot be equal to 0

In summary, the range of values for m is 0

Solution: It can be obtained from the question.

0＜|m|-1＜2

1＜|m|＜3

3 m -1, or 1 m 3

i.e. m (-3,-1) (1,3).

The focus is on the ellipse on the y-axis.

then 3-m>1-m>0

3-m>1-m constant was established.

So 1-m>0

m<1

For an ellipse, x 2 m+y 2 (1-m)=1

m>0,1-m>0

Focus y-axis. then 1-m>m>0

So 0