What is the equation of the line tangent to theĀ Ellipse 4x^2 + 2y^2 = 9 and which passes through the point (2, 6)?

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

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`4x^2 + 2y^2 = 9`

This is NOT a circle, This is an equation of an Ellipse.

We will find the derivative.

==> `8x + 4yy' = 0`

`==> y' = (-8x)/(4y)= (-2x)/y = m`

`==> y-y1 = m (x-x1)`

`==> y-6 = (-2x/y) ( x-2)`

`==> y^2 - 6y = -2x(x-2)`

`==> y^2 - 6y = -2x^2 + 4x`

```==> y^2 + 2x^2 - 6y - 4x = 0`

Now we will multiply (2) by -2 and add to (1)

==> `12y + 8x = 9 `

`==> y= (9-8x)/12`

`` Now we will substitute (3) into (1).

==>`2y^2 + 4x^2 = 9 `

`==> 2((9-8x)/12)^2 + 4x^2 = 9`

`==> 2( 81 - 144x + 64x^2)/144 + 4x^2 = 9`

`==> (81 - 144x + 64x^2)/72 + 4x^2 = 9`

`==> 81 - 144x + 64x^2 + 4x^2 =648`

`==> 68x^2 - 144x -567 = 0`

`==> x1= (144 +418.3)/136 = 4.13==> y1=(9-8(4.13))/12 = -2`

`==> x2= (144-418.3)/136 = -2.02 ==> y2= (9-8(-2.02))/12 = 2.1`

Then, there are 2 tangent lines at the points (4.13, -2) and ( -2.02, 2.1).

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