A Question about Tangent of the ellipse

[TheWizardMe]

Hi all how is everyone hope good and fine

i am new here but really a good forum which share very very big and important knowlegde

a small problem face me in get a solution for the question:

-if tangent line to the ellipse 9x^2 + 4y^2 = 36 has y intercept 6

find the equation of the tangent.

the problem is when we substitute y by 6 and try to get a value for x the equation will be as follow x^2 = -12 the problem is there is no roots for a negative number otherwise we use an imaginry numbers

hope anyone could help me to a get a solution for that question

thanks in advance

[Tom Mattson]

the problem is when we substitute y by 6 and try to get a value for x the equation (snip)

 

You shouldn't be trying that at all. When the problem states that the y-intercept of the tangent line is at [imath](0' date='6)[/imath'], you are not being told that [imath]y=6[/imath] is in the range of the ellipse.

 

You don't know where the tangent line intersects the ellipse, so call the point something like [imath](x_0,y_0)[/imath]. From there you can find the slope of the line in terms of the unknown point.

[TheWizardMe]

thanks Mr. Tom for replying me but i wanna more disscus for it cause i didnt got U

the question says that y intercept 6 so U mean that it intercept the line at (0,6) i think that this also means the f(x) which is y is equal 6 so i can subsititue y by 6

so PLZ Mr.Tom clarify it more cause i am confused about this question

thanks in advance

[Tom Mattson]

i think that this also means the f(x) which is y is equal 6 so i can subsititue y by 6

 

Certainly not. Look at the equation of the ellipse:

 

[math]9x^2+4y^2=36[/math]

 

Put it in standard form (do this by dividing by 36):

 

[math]\frac{x^2}{4}+\frac{y^2}{9}=1[/math]

 

[math]\frac{x^2}{2^2}+\frac{y^2}{3^2}=1[/math]

 

As you can see from this equation' date=' the [b']highest[/b] point on the ellipse has a y-coordinate of 3. So y=6 is quite out of the question.