Thread: Tangent Passing Through a Point & Implicit Differentiation

1. Tangent Passing Through a Point & Implicit Differentiation

Question:
Find the equation(s) of the tangents to $x^2 + 4y^2 = 16$ which pass through
a) the point (2,2)
b) the point (4,6).

Attempted Solution:
$y = \pm\sqrt{\frac{-x^2 + 16}{4}}$

$y' = \frac{-x}{4y}$

a) For $P(2,2)$:

Let $P_{tan}$ be parametrized as $(a, f(a)$.

$\frac{2 - \pm\sqrt{\frac{-a^2 + 16}{4}}}{2 - a} = \frac{rise}{run}= m$

What do I do from here? The points given are not points of tangency; they are points that the tangents pass through. So, do I have to set something equal and solve for $a$?

2. $\frac{y-2}{x-2}=[\frac{dy}{dx}]_{(x,y)=(2,2)}$

3. Do I have to sub P(2,2) into the derivative? What exactly am I trying to solve for?

4. after gettin $\frac{dy}{dx}$, substitute x by 2 and y by 2.

5. That leaves me with a slope of 1/4. Do I set 1/4 equal to the rise/run equation in my OP?

6. That leaves you with the equation $\frac{y-2}{x-2}=1/4$

7. Originally Posted by Sambit
$\frac{y-2}{x-2}=[\frac{dy}{dx}]_{(x,y)=(2,2)}$
We need to take care as the point $(2,2)$ is not on the ellipse.

If you plot the question both answers can be read off the graph. The tangent lines must be horizontal and vertical. Note that

$\displaystyle x^2+4y^2=16 \iff \frac{x^2}{4^2}+\frac{y^2}{2^2}=1$

Is the equation in standard form.

To solve it algebraically we need to find a point $(a,b)$ on the ellipse e.g it must satisfy $a^2+4b^2=16$ and

$\frac{b-2}{a-2}=\frac{dy}{dx}\bigg|(a,b)=-\frac{a}{4b}$

Multiplying out the bottom equation gives

$4b^2+a^2=2a+8b$ but since it must be on the ellipse $a^2+4b^2=16$ this gives

$16=2a+8b \implies a=8-4b=4(2-b)$

Now putting this back into the ellipse again gives

$[4(2-b)]^2+4b^2=16 \iff 5b^2-16b+12=0 \iff (5b-6)(b-2)=0$ as you can check this gives $b=2 \implies a=0$ and correctly gives the point $(0,2)$ as when one of the tangent lines occur.