Find the coordinates of the intersection of the following lines and ellipses

1.By the elliptic equation, the general divide, 4x 2 + 25y 2 = 100 i.e. 16x 2 + 100y 2 = 400

By the equation of the straight line, 10y=25-3x so 100y2=(25-3x) 2, substituting the above equation, 16x 2+(25-3x) 2 = 400 x=3 y=8 5 intersection points (3,2.).From the previous equation, x 2=y-2, substitute the latter equation, and solve y=2, -9 4 (rounded).

Intersection (0,2).

1) 3x+10y-25=0,x^2/25+y^2/4=1 ;

Result: x=3, y=8 5

2) x 2-y+2=0, x 2 16+y 2 4=1 Result: x1=0, y1=2

There is also a pair of complex solutions: x2=1 2 17*i; x3=-1/2√17*i; y2,3=-9/4

Choose plural solutions according to your own needs.

What does x 2 25 mean? Please make it clear in words!

<> "The elliptic equation sought by the meaning of the question is <>

<> and <>< 3 points.

Therefore, the standard equation for the ellipse is <>

……6 points. It is <> from the meaning of the title

<> about straight line <>

The symmetry points of 0 are:

<> let the standard equation of the hyperbola be <>

The half-focal length <> is known from the title

Therefore<> the standard equation for the hyperbola is <>

…13 points.

1.By the elliptic equation, the general division.

4x^2+25y^2

That is, 16x 2+100y 2

Equations from straight lines.

10y=25-3x

So 100y 2=(25-3x) 2, substituting the above equation, 16x 2+(25-3x) 2

x=3y=8/5

The intersection point (3, from the previous equation, x 2=y-2, is substituted by the latter isosensitive modification, and the intersection is solved by y=2, -9 4 (rounded).

The standard equation for an ellipse is y281+x29 1, a=9, b=3, c=62

Focus coordinates: Dansun stupid (0, 62), vertex coordinates (0, 9), (3, 0), eccentricity e=ca=223 modulus.

It is known that the focal points of the ellipse are and , and the length of the long axis is , and the ellipse of the line intersection is set at two points, and the coordinates of the midpoint of the line segment are found. The midpoint coordinates of the line segment are Let the equation of the ellipse macro orange movement be: , from the meaning of the title: , so the equation of the ellipse is the shield: , from , because of the difference of the quadratic equation, so the line and the ellipse have two different intersection points, let , then , so the coordinates of the midpoint of the line segment are .

Consider setting the focus to the right focus.

First of all, if the slope of l does not exist, i.e., l is perpendicular to the x-axis, then the coordinates of m,n can be found to be (c,b2 a)(c,-b2 a) a,b coordinates are (a,0)(-a,0) straight line am, bn is y=b2(x-a) (ac-a2), y=-b2(x+a) (ac+a2) x=a2 c can be solved by a synthesis

If the slope exists and is set to k, the straight line l is y=k(x-c) and the ellipse is synapsual(b2+a2k2)x2-2 (ca2k2)x+a2c2k2-a2b2=0 from Vedica's theorem, which can be expressed x1+x2 , x1*x2 , and then x1-x2 Next, let m (then the line am:y=y1(x-a) (x1-a) bn:y=y2(x+a) (x2+a) The abscissa of the intersection point of the two lines is [a(x1y2+x2y1)+a2(yi-y2)] x2y1-x1y2+a(y1+y2)] You may not be able to do it here, but don't forget the condition y1=k(x1-c) y2=k(x2-c) The expression of y1 y2 is left with x1+x2 , x1*x2 , x1-x2 and then they are solved by Vedic theorem by substituting them with a,b,c,k The specific formula is too troublesome, don't play, You should be able to make it after reading it.

Straight line x+3y-6=0

Let y=0, x=6

Let x=0, y=2

If b = 2, c = 6

a²=b²+c²=40

The ellipse is a clear equation: x 40 + y 4 = 1 if c = 2, b = 6

Then a = b + c = 40

Elliptic equation: y refers to 40+x 36=0 before rolling