find the values of c if a striaght line y = 2x + c, is tangent to the ellipse:
(x^2)/9 + (y^2)/4 = 1
Using the top half of the ellipse, we have
y = 2/3 √(9-x^2)
For the ellipse to intersect the line y=2x+c, we need
2/3 √(9-x^2) = 2x+c
4/9 (9-x^2) = (2x+c)^2
40/9 x^2 + 4cx + c^2-4
For the ellipse to be tangent to the line (only one solution), we need the discriminant to be zero.
16c^2 - 4(40/9)(c^2-4) = 0
So, c = 2√10
See the graph at
https://www.wolframalpha.com/input/?i=plot+(x%5E2)%2F9+%2B+(y%5E2)%2F4+%3D+1,+y%3D2x%2B2%E2%88%9A10
There is a similar point where c = -2√10
To find the values of c for which the straight line y = 2x + c is tangent to the ellipse, we need to find the points of tangency between the line and the ellipse.
The equation of the given ellipse is (x^2)/9 + (y^2)/4 = 1.
First, let's substitute the equation of the line into the equation of the ellipse:
(x^2)/9 + ((2x + c)^2)/4 = 1
Simplifying this equation, expand the square of (2x + c):
(x^2)/9 + (4x^2 + 4cx + c^2)/4 = 1
Next, let's multiply the entire equation by 36 to eliminate the denominators:
4x^2 + 9(4x^2 + 4cx + c^2) = 36
Expand and simplify further:
4x^2 + 36x^2 + 36cx + 9c^2 = 36
Combine like terms:
40x^2 + 36cx + 9c^2 = 36
To determine the points of tangency, we need this quadratic equation to have only one solution, which occurs when the discriminant is zero.
The discriminant of the quadratic equation ax^2 + bx + c = 0 is given by b^2 - 4ac.
In our case, the discriminant is:
(36c)^2 - 4(40)(9c^2)
Simplifying this equation:
1296c^2 - 1440c^2 = 0
Subtracting the terms:
-144c^2 = 0
Dividing both sides by -144:
c^2 = 0
Taking the square root of both sides gives:
c = 0
Therefore, the value of c for which the straight line y = 2x + c is tangent to the ellipse (x^2)/9 + (y^2)/4 = 1 is c = 0.