I have a right circular cylinder (circle on the x-z axes) and a plane on the x-z axes. If the plane is tilted 40 degrees from the y-axis to the cylinder, is there a
straight and direct (ie simple) solution to find the equation on the curve where the plane intersects the cylinder?
Obviously if the plane was 0 degrees to the axis of the cylinder, the equation would be the equation in the circular form of the circle (x square + y square = constant). But I do not remember from the weak past when I took analytical geometry * how to put the problem together in the first place. In part, it's because I can't remember (nor does Google help me at the moment) how I enter the plane's equation.
Any helpful suggestions, even pointing out a web page that spell my problem out in small words and short sentences, would be appreciated.
* I took "trigonometry and analytical geometry" in the first semester of my senior year of high school. It was in 1976. I haven't used neither trig nor analytical geometry since then …