20 12 national college entrance examination mathematics new curriculum standard 20 titled 2P what is BD?
Look at the problem, it is not an equation that requires a circle, so let r be the radius. Look at BF=FD=R, and know that the angle is 90. Then the square of BD is equal to twice the square of R. How to find BD? First, the abscissa of the directrix is known. If you substitute it into the equation of a circle with f as the center, it is set. Then, you can find out that Y and BD are twice as much as Y, and then substitute them into the above trigonometric equation to get the relationship between R and P. Then BD can also be expressed by P. When you do analytic geometry, you must follow the topic and start with the conditions, and you can't take it for granted. F is the center of the circle, a is not.