Find the equation of the line joining the points of contact of the tangents to a curve

Question

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 $3x+4y=0$

Answer 1

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.

Answer 2

Let $f(x,y)=x^2+2y^2$. Then $\nabla f(x,y)=(2x,4y)$. For which points $(x,y)$ from that curve is it true that $\nabla f(x,y)$ is orthogonal to the line $2x-3y=c$? That occurs if and only if$$\left\{\begin{array}{l}\frac{4y}{2x}=-\frac32\\x^2+2y^2=2.\end{array}\right.$$This system has two solutions: $\pm\left(-\frac4{\sqrt{17}},\frac3{\sqrt{17}}\right)$. Besides,$$2\times\left(-\frac4{\sqrt{17}}\right)-3\times\frac3{\sqrt{17}}=-\sqrt{17}\quad\text{and}\quad2\times\left(\frac4{\sqrt{17}}\right)-3\times\left(-\frac3{\sqrt{17}}\right)=\sqrt{17},$$and therefore the possible values of $c$ are $\pm\sqrt{17}(\approx\pm4.12)$.

The line joining the points of contact is the line joining $\left(-\frac4{\sqrt{17}},\frac3{\sqrt{17}}\right)$ and $\left(\frac4{\sqrt{17}},-\frac3{\sqrt{17}}\right)$, which is $3x+4y=0$.

Answer 3

The equation of ellipse is $\displaystyle \frac{x^2}{2} + \frac{y^2}{1} = 1 \implies a = \sqrt2, b = 1$ where $a$ is semi-major axis and $b$ is semi-minor axis.

If you have already found value of $c = \pm \sqrt{17}$, you know the equations of tangent lines are,

$ \displaystyle 2x - 3y = \pm \sqrt{17} \implies \pm\frac{2x}{\sqrt{17}} \mp \frac{3y}{\sqrt{17}} = 1$ ...$(i)$

We know that tangent to a point $(x_1, y_1)$ on ellipse $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is given by $\displaystyle \frac{x x_1}{a^2} + \frac{y y_1}{b^2} = 1$

Comparing with $(i)$, $\displaystyle x_1 = \pm \frac{2a^2}{\sqrt{17}}, y = \mp \frac{3b^2}{\sqrt{17}}$. So points are $ \displaystyle \left(\pm \frac{4}{\sqrt{17}}, \mp \frac{3}{\sqrt{17}} \right)$.

The line between the two points is indeed $3x + 4y = 0$