find the equation of the tangents to a quadratic curve through the point $(-7,1)$

Question

Given a quadratic curve $\mathcal{C}: X^2+2XY+4Y^2-2X+6Y+6=0$, determine its type and find the equation of the tangents to $\mathcal{C}$ through the point $(-7,1)$.

So far I've written the matrix of $\mathcal{C}$ as $$A = \begin{bmatrix}6 & -1 & 3 \\ -1 & 1 & 1 \\ 3 &1 &4\end{bmatrix}.$$

I used the fact that $\det A\neq 0$ and $\begin{vmatrix}1 & 1 \\ 1 & 4\end{vmatrix}>0$ to say that this is an ellipse, by some theorems we did in class. Now I know that if a point $P(x,y)$ is on the curve, then the unique tangent at this point is the polar line which would be calculated by taking the homogeneous vector of $P$ and multiplying it by A as follows $(1:x:y)A$. However $(-7,1)$ is not on the curve although I calculated $x=\frac{16}{7}$ to be its tangent if it was. so I am stuck here. What could I use to find the tangents since the point is not on the curve? If we write $t_P: y=ax+b$ then $t_P$ must contain $P$ and intersect $\mathcal{C}$ at a point, but I am not sure what point that would be.

Answer 1

The first example has been treated.

Let us consider the second example you have given in one of your comments:

$$x^2-xy-y^2-2x+2y+1=0 \tag{1}$$

(which is a hyperbola: see figure below) with point $P(4,-2)$.

The generic parametric equations of lines passing through point $P$ are:

$$\begin{cases}x&=& \ \ \ 4+at\\y&=&-2+bt\end{cases}$$

If we substitute them in (1), we obtain the following quadratic equation in variable $t$:

$$(a^2 - ab - b^2)t^2 + 2(4a + b)t + 9=0$$

Its discriminant is, up to an unimportant factor $4$, equal to :

$$\Delta=(4a + b)^2-9(a^2 - ab - b^2)$$

$$\Delta=7a^2 + 17ab + 10b^2\tag{2}$$

Setting $\Delta=0$ is expressing that there is a double root, a fact characterizing a tangent line.

Let us take arbitrarily $a=1$ in (2). The two possible roots of quadratic equation obtained from (2)

$$7 + 17b + 10b^2=0$$

are $b=-1$ and $b=-0.7$ ($b$ is in fact plainly the slope of the straight lines)

Therefore the two parametric equations of tangent lines issued from point $P(4,-2)$ are:

$$\begin{cases}x&=& \ \ \ 4+t\\y&=&-2-t\end{cases} \ \text{and} \ \begin{cases}x&=&\ \ \ 4+t\\y&=&-2-0.7t\end{cases}$$

otherwise said, with resp. cartesian equations:

$$x+y=2 \ \ \ \ \text{and} \ \ \ \ 7x+10y=8$$

Answer 2

An approach, which contains some elements of calculus, is considered here to find the tangents to the quadadratic curve, the equation of which is given as $$f \left(x,y\right) =x^2-xy-y^2-2x+2y+1=0. \tag{1}$$

We start by differentiating the equation as shown below to obtain the slope of the tangent to the curve (1) at an arbitrary point $Q\left(x,\space y\right)$ lying on it $$2xdx-ydx-xdy-2ydy-2dx+2dy=0\quad\rightarrow\quad \frac{dy}{dx}= \frac{2x-y-2}{ x+2y-2}. \tag{2}$$

If this tangent passes through a point $P \left(4, -2\right)$, which, as we know, does not lie on the given curve, we can obtain a second expression to describe its slope as, $$\frac{dy}{dx}=\frac{y+2}{x-4}. \tag{3}$$

We can equate (3) to (2) to establish the following quadratic relationship between $x$ and $y$. $$\frac{2x-y-2}{ x+2y-2}= \frac{y+2}{x-4}\quad\rightarrow\quad x^2-xy-y^2-6x+y+6=0 \tag{4}$$

Now, we strive to express $y$ in terms of $x$ as simple as possible. In order to obtain a linear relationship between $x$ and $y$, we need to subtract (4) from (1). The result is shown below. $$4x-y-5=0 \quad\rightarrow\quad y=5-4x \tag{5}$$

Next, we seek for an equation with one variable, so that we can solve for that variable to determine either $x-$ or $y-$coordinate of the point $Q$. If we proceed by replacing $y$ in (1) using its value given by (5), we end up with the following quadratic equation of a single variable. $$11x^2 - 25x+14=0 \quad\rightarrow\quad \left(x-1\right) \left(11x-14\right) = 0 \tag{6}$$

This equation has two real roots – meaning that there are two tangents that go through the point $P$. The roots of (6) are $$x_1=1 \quad\mathrm{and}\quad x_2=\frac{14}{11}. \tag{7}$$

Now, we can use (3) to find the slope of each tangent. But, first we have to substitute the value given by (5) in (3) to obtain $$\frac{dy}{dx}=\frac{y+2}{x-4}=\frac{7-4x}{x-4}. \tag{8} $$

According to (7) and (8), the slopes are $$ \left(\frac{dy}{dx} \right)_1 = -1 \quad\mathrm{and}\quad \left(\frac{dy}{dx} \right)_2 =-\frac{7}{10}.$$

Therefore, the equations of the two tangents can be written as $$y+x+c_1=0 \quad\mathrm{and}\quad y+\frac{7}{10}x+c_2=0.$$

Since the point $P \left(4, -2\right)$ sits on both these tangents, we can find the $y$-intercepts $c_1$ and $c_2$ as $$c_1=-2 \quad\mathrm{and}\quad c_2=-\frac{4}{5}.$$