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
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.
To prove that the line x-2y+10=0 touches the ellipse 9x^2+64y^2=576, we can find the point of intersection between the line and the ellipse. If the line intersects the ellipse at a single point, it means it touches the ellipse.
Step 1: Find the coordinates of the point of intersection.
Substitute y = (x+10)/2 into the equation of the ellipse:
9x^2 + 64((x+10)/2)^2 = 576
Simplifying the equation:
9x^2 + 64(x+10)^2/4 = 576
36x^2 + 64(x+10)^2 = 2304
36x^2 + 64(x^2 + 20x + 100) = 2304
36x^2 + 64x^2 + 1280x + 6400 = 2304
100x^2 + 1280x + 6400 - 2304 = 0
100x^2 + 1280x + 4096 = 0
Step 2: Solve the quadratic equation.
Use the quadratic formula to solve for x:
x = (-b ± √(b^2 - 4ac))/(2a)
where a = 100, b = 1280, and c = 4096.
Calculating the values:
x = (-1280 ± √(1280^2 - 4*100*4096))/(2*100)
x = (-1280 ± √(1638400 - 1638400))/(200)
x = -1280/200
x = -6.4
Step 3: Substitute the value of x back into the line equation to find y.
Using the equation x - 2y + 10 = 0:
-6.4 - 2y + 10 = 0
-2y = 6.4 - 10
-2y = -3.6
y = -3.6/-2
y = 1.8
Therefore, the point of intersection is (-6.4, 1.8). Since the line intersects the ellipse at a single point, we can conclude that the line x-2y+10=0 touches the ellipse 9x^2+64y^2=576.