SOLUTION: Find the point on the ellipse 𝑥^2 + 5𝑦^2 = 36 which is the closest, and which is the farthest point from the line 2𝑥 − 5𝑦 + 30 = 0. Sketch the graph

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Find the point on the ellipse 𝑥^2 + 5𝑦^2 = 36 which is the closest, and which is the farthest point from the line 2𝑥 − 5𝑦 + 30 = 0. Sketch the graph      Log On


   

Question 1200122: Find the point on the ellipse 𝑥^2 + 5𝑦^2 = 36 which is the closest, and which is the farthest point from the line 2𝑥 − 5𝑦 + 30 = 0. Sketch the graph
Found 2 solutions by Edwin McCravy, mccravyedwin:
Answer by Edwin McCravy(20059) About Me  (Show Source):
You can put this solution on YOUR website!


We draw two lines tangent to the ellipse parallel to the given line.
Their point of tangency will be the closest and farthest points
on the ellipse from the given line. 



x%5E2+%2B+5y%5E2+=+36

Differentiate implicitly:

2x+%2B+10y%2Aexpr%28dy%2Fdx%29=0
Solve for expr%28dy%2Fdx%29

dy%2Fdx=-x%2F%285y%29

We find the slope of the line:

2x+%E2%88%92+5y+%2B+30+=+0
y=expr%282%2F5%29x%2B6
slope=2%2F5

Find the points on the ellipse where the slope of the tangent lines
are parallel to the given line by setting the derivative of ellipse
equal to slope of line: 

x%2F%285y%29=2%2F5
solve for x
x=-2y 

Substitute in ellipse's equation

x%5E2+%2B+5y%5E2+=+36
%28-2y%29%5E2%2B5y%5E2=36
4y%5E2%2B5y%5E2=36
9y%5E2=36
y%5E2=4
y=%22%22+%2B-+2

x%5E2+%2B+5y%5E2+=+36
x%5E2+%2B+5%28%22%22+%2B-+2%29%5E2=36
x%5E2+%2B+5%284%29=36
x%5E2+%2B20=36
x%5E2=16
x=%22%22+%2B-+4

Four points on the ellipse are (-4,2), (-4,-2), (4,2), (4,-2)

The only 2 points on the green tangent lines above are (-4,2) and (4,-2).

They are the required points.

Edwin

Answer by mccravyedwin(407) About Me  (Show Source):
You can put this solution on YOUR website!
We can also do the problem without using calculus:



We draw two lines tangent to the ellipse parallel to the given line.
Their points of tangency will be the closest and farthest points
on the ellipse from the given line. 



We put the given line in slope-intercept form

2x+%E2%88%92+5y+%2B+30+=+0
-5y=-2x-30
%28-5y%29%2F%28-5y%29=expr%28%28-2%29%2F%28-5%29%29x-30%2F%28-5%29
y=expr%282%2F5%29x%2B6

The green tangent lines will have the same slope with a different
y-intercept b

So either green tangent line will have equation y=expr%282%2F5%29x%2Bb

When we solve the system by substitution

system%28x%5E2%2B5y%5E2=36%2Cy=expr%282%2F5%29x%2Bb%29

There must be one and only one point of intersection.

x%5E2%2B5%28expr%282%2F5%29x%2Bb%29%5E2=36%29
x%5E2%2B5%28expr%284%2F25%29x%5E2%2Bexpr%284%2F5%29bx%2Bb%5E2%29=36%29
x%5E2%2Bexpr%284%2F5%29x%5E2%2B4bx%2B5b%5E2=36%29
5x%5E2%2B4x%5E2%2B20bx%2B25b%5E2=180%29
9x%5E2%2B20bx%2B25b%5E2-180=0%29

For there to be only one solution for the point of tangency, the
discriminant must = 0
Discriminant+=+%2820b%29%5E2-4%289%29%2825b%5E2-180%29=0
400b%5E2-4%289%29%2825b%5E2-180%29=0
100b%5E2-9%2825b%5E2-180%29=0
100b%5E2-225b%5E2%2B1620=0
-125b%5E2%2B1620=0
-125b%5E2=-1620
b%5E2=expr%28%28-1620%29%2F%28-125%29%29%0D%0A%7B%7B%7Bb%5E2=324%2F25
b=+%22%22+%2B-+18%2F5

Substituting b=18%2F5
9x%5E2%2B20bx%2B25b%5E2-180=0%29
9x%5E2%2B20%2818%2F5%29x%2B25%2818%2F5%29%5E2-180=0%29
9x%5E2%2B72x%2B324-180=0%29
9x%5E2%2B72x%2B144=0%29
x%5E2%2B8x%2B16=0
%28x%2B4%29%28x%2B4%29=0
x=-4
Substituting in the equation of the line:
y=expr%282%2F5%29x%2Bb%29
y=%282%2F5%29%28-4%29%2B18%2F5%29
y=2

So one point is (-4,2), that's the closest point.

Substituting b=-18%2F5
9x%5E2%2B20bx%2B25b%5E2-180=0%29
9x%5E2%2B20%28-18%2F5%29x%2B25%2818%2F5%29%5E2-180=0%29
9x%5E2-72x%2B324-180=0%29
9x%5E2-72x%2B144=0%29
x%5E2-8x%2B16=0
%28x-4%29%28x-4%29=0
x=4
Substituting in the equation of the line:
y=expr%282%2F5%29x%2Bb%29
y=%282%2F5%29%284%29-18%2F5%29
y=-2

So the other point is (4,-2), that's the farthest point.

Edwin