# Define the equations of the tangents of the ellipse 4x^2+5y^2=120 perpendicular to the line x+2y+13=0.

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### 1 Answer

The equations of the tangents of the ellipse 4x^2+5y^2=120 perpendicular to the line x+2y+13=0 have to be determined.

x + 2y + 13 = 0

=> `y = -x/2 - 13/2`

The slope of this line is `-1/2` . The slope of lines perpendicular to the line is 2.

The slope of the tangent to the curve at any point `4x^2+5y^2=120` is the value of `dy/dx` at that point. Using implicit differentiation

`8x + 10y*(dy/dx) = 0`

=> `dy/dx = (-8x)/(10y) = -(4/5)*(x/y)`

Equating this to 2, `-(4/5)*(x/y) = 2`

=> `x/y = -5/2`

=> `x = (-5/2)*y`

substitute this in 4x^2+5y^2=120

=> 4*(25/4)*y^2 + 5y^2 = 120

=> 25y^2 + 5y^2 = 120

=> 30y^2 = 120

=> y^2 = 4

=> y = 2 and y = -2

=> x = -5 and x = 5

The equation of the tangents are: (y - 2)/(x + 5) = 2

=> y - 2 = 2x + 10

=> 2x - y + 12 = 0

and (y + 2)/(x - 5) = 2

=> y + 2 = 2x - 10

=> 2x - y - 12 = 0

In the graph drawn below the line in green is the reference line given and the lines in red are the tangents.

**The required equation of the tangents are 2x - y + 12 = 0 and 2x - y - 12 = 0.**