Considering the intersection of straight line with ellipse and hyperbola, it is usually necessary to combine straight line equation with ellipse or hyperbola equation.

If a straight line is juxtaposed with an ellipse, let's set a straight line: Ax+By+C=0 and an ellipse: x 2/a 2+y 2/b 2 =1.

Then the equation obtained at the same time is:

(a^2*a^2+b^2*b^2)x^2+2aca^2*x+a^2*(c^2-b^2*b^2)=0

(a^2*a^2+b^2*b^2)y^2+2bcb^2*y+b^2*(c^2-a^2*a^2)=0

Just remember the above formula, and then a lot of things will be easy to ask. (Push yourself, remember more clearly)

For example, if the straight line AB intersects the ellipse at point A and point B, what conditions will you give to prove that OA is perpendicular to OB? If the linear equation is set to y=kx+m instead of Ax+By+C=0, only the equation about x will be obtained, but x 1x2+y 1y2 will be expressed as x1x2+(kx1+m) (kx2+).

This method can also be used to find the length of straight line AB and the area of triangle OAB, and even more convenient formulas can be derived.

To calculate a hyperbola whose focus is on the X axis, you only need to change all B 2 in the simultaneous equations into -B 2, and you will know it on the Y axis.

However, this method also has some problems that are difficult to solve. You can try to use x=x0+t*cos a, y=y0+t*sin a (t is the parameter and a is the inclination angle) to solve those problems (point angle parameter equation).