For the first part, require that
x = 2y -10 . Then insert that value of x in the second (ellipse) equation and see if a solution exists. If it does, there is a point that satisfies both equations.
Part 2, determine the slope dy/dx of the tangent line at (x = a sin theta, y = b cos theta)
2x/a^2 + (2y/b^2) dy/dx = 0
dy/dx = -(x/y)(b/a)^2
= (tan theta)(b/a)
(theta is measured counterclockwise from the +x axis)
Knowing the slope and the given coordinates (x,y) of one point on the tangent line, the equation for the tangent line can be written.
conic sections
prove that the line x-2y+10=0 touches the ellipse 9x^2+64y^2=576
show that the equation of the normal to the ellipse x^2/a^2 + y^2/b^2 = 1 at the point (a cos theta, b sin theta ) is ax sin theta - by cos theta - (a^2-b^2) sin theta cos theta =0
1 answer
1 answer