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I am looking to find the equation of the line tangent to


at the point $(0,-2)$.

I have a feeling I need to implicitly differentiate here?

Am I on the right track?

What do I do after finding $y'$ to actually find the solution? Like what steps do I take to find the tangent line?

Not asking for the solution but a push in the right direction would be helpful, although a solution would be nice to look over.

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You don't absolutely need implicit differentiation, since we can solve explicitly for the suitable $y$. But implicit is far easier. Can do it with no paper in a second or two. –  André Nicolas Jun 29 '12 at 23:45

3 Answers 3

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Yes you do need to use implicit differentiation. When finding a tangent line, you nearly always need the point-slope formula:

$$y_2 - y_1 = m(x_2 - x_1)$$

Solution to the implicit differentiation is below (to check your work). Simply hover your mouse over the grey box.

$y^2(y^2 - 4) = x^2(x^2 -5) \\\\$ Multiplying the polynomials gets us to $y^4 - 4y^2 = x^4 - 5x^2$. Taking the derivative with respect to $x$ gets us: $4y^3y' - >! 8yy'=4x^3 - 10x$. Factoring to get $y'$ by itself: $y'(4y^3 - 8y) = 4x^3 - 10)$. Divide through to get $y'$ by itself: $y' = \dfrac{4x^3 - 10x}{4y^3-8y}$. You could make your life a bit easier by factoring this into $y' = \dfrac{2x(2x^2 - 5)}{4y(y^2-2)}$. You could cancel out a factor of $2$ to get $y' = \dfrac{x(2x^2 - 5)}{2y(y^2-2)}$. To find the slope, plug in your points $x = 0, y = -2$ into our equation for $y'$ to find the slope of the line. Note that the slope is $0$. To find the equation of the tangent line, use that value for $m$ you just found ($m=0$) and your given points into the point-slope formula and you find that the tangent line is $y=-2$.


Our expression for $y'$ is:

$$y' = \dfrac{2x(2x^2 - 5)}{4y(y^2-2)}$$

We were given a point $(0,-2)$. So, $x = 0, y = -2.$ Plugging this into the expression above yields:

$$y' = 0.$$

So, our slope of the line tangent to the curve $y^4-4y^2 = x^4-5x^2$ is zero. Now, using the point-slope formula, we have:

$$y - -2 = 0(x-0)$$

$$y+2 = 0 \implies y = -2$$

So, the tangent line is simply the horizontal line $y=-2.$

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Working on it now lets see if i get it right –  soniccool Jun 29 '12 at 23:39
I think i found a problem you have $2x(x^2-5)$ in order to be factored from $4x^3-10x$ shouldnt that be $2x(2x^2-5)$ ? –  soniccool Jun 29 '12 at 23:40
Yup, you are correct. Simple slip of mind. –  Joe Jun 29 '12 at 23:46
:) Awesome so basically would this suffice as an answer for the equation of the tangent line? $$(y-2)=(\frac{x(x^2-5)}{2(y^2-2)}) (x-0) $$ Or would i have to solve from there? –  soniccool Jun 29 '12 at 23:48
Ah i totally forgot about that Alright awesome! :D –  soniccool Jun 29 '12 at 23:56


Yes, you need to implicitly differentiate. In order to describe a line, you need a point and a slope. You already have the point, so you just need to extract the slope from the implicit differentiation.

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Awesome okay let me try this –  soniccool Jun 29 '12 at 23:04

Differentiate both sides with respect to $x$. On the right, we get $4x^3-10x$. On the left we get $(4y^3-8y)\frac{dy}{dx}$.

Now (crucial) we need to evaluate the derivative at $(0,-2)$. At this point, $4x^3-10x$ is $0$, and $4y^3-8y$ is not. So $\left(\frac{dy}{dx}\right)_{(0,-2)}=0$.

Slope is $0$, we pass through $(0,-2)$, so the tangent line is horizontal, and has equation $y=-2$.

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