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Find the equations of the vertical and horizontal tangent lines of the following ellipse.

16x2 + y2 - 128x + 3y + 256 = 0

User Ieatczp
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1 Answer

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Answer:

  • x = 3.625; x = 4.375
  • y = 0; y = -3

Explanation:

We can rewrite the equation to standard form to see the center and semi-axis lengths.

16x^2 -128x +y^2 +3y = -256

16(x^2 -8x +16) +(y^2 +3y +2.25) = -256 +256 +2.25

16(x -4)^2 +(y +1.5)^2 = 9/4 . . . . . write as squares

(x -4)^2/(9/64) +(y +1.5)^2/(9/4) = 1 . . . . divide by 9/4

((x -4)/0.375)^2 +((y +1.5)/1.5)^2 = 1 . . . . put in useful form

In this form, we have ...

((x -h)/a)^2 +((y -k)/b)^2 = 1

where (h, k) is the center, 2a is the length of the axis in the x-direction, and 2b is the length of the axis in the y-direction. The required tangents are ...

x = h±a

y = k±b

For the given ellipse, the tangent lines are ...

  • x = 4 -0.375 = 3.625, x = 4.375
  • y = -1.5 -1.5 = -3, y = -1.5 +1.5 = 0
Find the equations of the vertical and horizontal tangent lines of the following ellipse-example-1
User RonakThakkar
by
5.0k points