I study maths as a hobby. I am stuck on this question:
Find the values of c for which the line $2x-3y = c$ is a tangent to the curve $x^2+2y^2=2$ and find the equation of the line joining the points of contact.
I have established that $c=\pm \sqrt17$.
To get the line joining the points of contact I thought one way would be to start by finding the points of contact.
For $c =\sqrt 17$ the equation of the line is y = $\frac{2x-\sqrt 17}{3}$ and the equation of the curve is $y = \sqrt \frac{2-x^2}{2}$
At the tangent
$\frac{2x-\sqrt 17}{3} = \sqrt \frac{2-x^2}{2}$
$\rightarrow \frac{4x^2-4x\sqrt17+17}{9} = \frac{2-x^2}{2}$
$\rightarrow 17x^2-8x\sqrt17+16=0$
$\rightarrow x = \frac{8\sqrt17\pm\sqrt1088-1088}{34}=\frac{8\sqrt17}{34}$ but proceeding in this fashion seems very messy and I don't see how it will lead to the answer given in the book as $3x+4y=0$
Best Answer
Here is a slightly more calc 1 approach that Jose's answer.
You know that the tangent line you're looking for has slope $\frac{2}{3}$, so we can take the derivative of your curve with respect to $x$ and plug in that information.
The derivative of your curve, using implicit differentiation, is
$$ 2x+4yy'=0 \implies y' = -\frac{x}{2y}. $$
Plugging in the desired slope gives that $y=\frac{-3}{4}x$. Since we have this relation between the points, we plug back into the equation for the curve itself and find that
$$ x^2 + 2\cdot\frac{9}{16}x^2 = 2 \implies x= \pm \frac{4}{\sqrt{17}}, y= \mp\frac{3}{\sqrt{17}}. $$
With these two solutions, plug into the equation of the line to find the possible values for $c$:
$$ c=2\cdot\pm \frac{4}{\sqrt{17}} - 3\cdot \mp \frac{3}{\sqrt{17}} = \pm \sqrt{17}. $$
To address your final question, we know that any two points on the plane determine a line and we can write down that line using point slope form:
$$ (y-\frac{3}{\sqrt{17}}) = -\frac{3}{4}(x + \frac{4}{\sqrt{17}}) \implies y= -\frac{3}{4}x, $$ the line you desired.