Find the equations for the lines tangent to the ellipse 4x^2+y^2=72 that are perpendicular to the line x+2y+3=0
To find the equations for the lines tangent to the ellipse that are perpendicular to the given line, we need to follow these steps:
Step 1: Find the slope of the given line.
The given line is x + 2y + 3 = 0. To find its slope, we need to rewrite it in slope-intercept form (y = mx + b), where m is the slope. Subtracting x from both sides, we have 2y = -x - 3. Now divide by 2 to isolate y, which gives y = (-1/2)x - (3/2). So, the slope of this line is -1/2.
Step 2: Find the point(s) where the lines are tangent to the ellipse.
Since the lines are tangent to the ellipse, they intersect the ellipse at exactly one point. To find this point, we need to solve the system of equations formed by the ellipse equation and the equation of the line perpendicular to the given line.
The equation of the given line has a slope of -1/2, so the line perpendicular to it will have a slope of 2 (since the product of slopes for perpendicular lines is -1).
Equation of the line perpendicular to the given line: y = 2x + b
Substituting this into the equation of the ellipse, we get:
4x^2 + (2x + b)^2 = 72
Expand and simplify:
4x^2 + 4x^2 + 4bx + b^2 = 72
8x^2 + 4bx + b^2 - 72 = 0
Step 3: Determine the conditions for tangency.
For the line to be tangent to the ellipse, the quadratic equation formed in Step 2 will have exactly one solution, which means its discriminant must be zero (b^2 - 4ac = 0).
So, in this case, the discriminant of the quadratic equation is:
4b^2 - 4(8)(b^2 - 72) = 0
16b^2 - 4(8b^2 - 576) = 0
16b^2 - 32b^2 + 2304 = 0
-16b^2 + 2304 = 0
Step 4: Solve for b.
Solving this quadratic equation, we get:
16b^2 = 2304
b^2 = 2304 / 16
b^2 = 144
b = ±12
So, we have two values for b: 12 and -12.
Step 5: Find the equations of the tangent lines.
Using the values of b, we can find the equations of the tangent lines by substituting them back into the equation of the line perpendicular to the given line:
Line 1: y = 2x + 12
Line 2: y = 2x - 12
Therefore, the equations for the lines tangent to the ellipse 4x^2 + y^2 = 72 that are perpendicular to the line x + 2y + 3 = 0 are:
Line 1: y = 2x + 12
Line 2: y = 2x - 12
If a line is perpendicular to
x+2y+3=0 (or y = -x/2 - 3/2),
the slope of the line must be 2.
Solve the equation dy/dx = 2.
y = sqrt(72 - 4x^2)
dy/dx = (-4x)/sqrt(72 - 4x^2) = 2