How to find the intersection of the line y=x+3 with the ellipse (x^2/9) + (y^2/4)=1?

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

To find a point of intersection you need to find the point (a,b) in which the line y=x+3 and the ellipse intesect.

first substitute y in the ellipse equation with x+3

==> (x^2/9) + y^2/4 = 1

by substituting y ,

==> (x^2 /9 + (x+3)^2/4 = 1   ,

==) 4x^2 + 9(x^2+6x+9) = 36

==> 4x^2 + 9x^2 + 54x+ 81 = 36

==> 13x^2 +54x + 81= 36

==> 13x^2 + 54x - 45 = 0

==> (13x + 15)(x+3) = 0

then x values: x1=  -15/13  andx2= -3

and y values y = x+3

then y1 = -15/13 + 3 = 24/13

and y2 = -3 + 3 = 0

then the line intersect with the ellipse in two points (-15/13, 24/13) and   (-3, 0)

==>

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neela | High School Teacher | (Level 3) Valedictorian

Posted on

Substitute y = x+3... (1)in the equation of the ellipse x^2/9+y^2/4 = 1........(2). Then we get y eliminated and the remaing equation is a second degree quadratic in x. Solve for x and substitute the value of x in y = x+3 to find the y value:

x^2/9+ (x+3)^2/4 = 1. Multiply by 9*4 both sides to get rid of the denominator:

4x^2+9(x+3)^2 = 36. Or

4x^2+9x^2+54x+81 = 36.

13x^2+54x-45 = 0

(13x+9)(x-5) = 0. Or

x=-9 /13 or x =5. From (1),  when x =-9/13,  y = -9/13+3 = -30/3 .

Also from (1) when x = 5, y = x+3 = 5+3 = 8.

So the points of intersection the line and the ellipse are ( -9,13 , -30/13) and (5 , 8)

Top Answer

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We know that the intersection of the graphs consists of the common points of the graphs, these common points being found by solving the equations of the graphs, simultaneously.

In our case, we'll substitute in the equation of the ellipse, the unknown y, by the expression from the equation of the line.

(x^2/9) + [(x+3)^2/4]=1

The common denominator of the 2 ratios is 36, so we'll multiply the first ratio by 4 and the second, by 9.

4x^2 + 9(x+3)^2=36

4x^2 + 9x^2 + 54x + 81 - 36 = 0

13x^2 + 54x + 45 = 0

Now we'll use the formula of the quadratic equations for finding the solutions.

x1 = [-54+sqrt(54^2 - 4*13*45)]/2*13

x1 = (-54 + sqrt(2916 - 2340))/26

x1 = (-54 + sqrt(576))/26

x1 = (-54+24)/26

x1 = -30/26

x1 = -15/13, so y1 = x1+3 = -15/13 + 3 = 24/13

x2 = (-54-24)/26

x2 = -78/26

x2 = -3, so y2 = -3+3=0

So, the intersection points are:

(-15/13, 24/13) and  (-3,0)

alic31wonder's profile pic

alic31wonder | Student, College Freshman | (Level 1) eNoter

Posted on

Nothing new in the second answer!

Perhaps the 3rd answer will be something else.

 

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