Finding equation of tangent lines at a point to the ellipse : confusion on passing point

Question

This is the question,

Equation of the tangent at the point (3, -1) to the ellipse $2x^2 + 9y^2 = 3$ is

Solution

my question is, shouldn't the tangent pass through the point given?

one of the option was $2x + 3y = 3$

And what if the question was like

Equation of the tangent from the point (3, -1) to the ellipse $2x^2 + 9y^2 = 3$ is

the only change is from the point can I still use this formula? and should the points satisfy the tangent in this case also?

Answer 1

Given the stardard form of the ellipse equation:

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ Suppose the tangent line touches the ellipse at point $B(x_1,y_1)$. First, we find the slope $k$ for the tangent line by taking the derivative of the ellipse equation,

$$\frac{2x_1}{a^2}+\frac{2y_1}{b^2} y'=0\Longrightarrow y'=k=-\frac{b^2x_1}{a^2y_1}$$

then we can get the equation for the tangent line

$$y-y_1=-\frac{b^2x_1}{a^2y_1}(x-x_1)$$

Simplify and we get:

$$\frac{x_1}{a^2}x+\frac{y_1}{b^2}y=1\tag{1}$$

The point $A(3, -1)$ is outside the ellipse $2x^2 + 9y^2 = 3$. So there are two lines passing point $A$ and tangent to the ellipse at point $B$.

Note this step is WRONG in your solution! First, the tangent point $B$ is currently unknown for us, and we need to find its coordinate $(x_1, y_1)$ later. We are given the point $A(3, -1)$ is on the tangent line, so we should plug in $x=3, y=-1$ and $a^2=\frac32, b^2=\frac13$ into eq.(1), hence we get

$$\frac{x_1}{\frac32}\cdot3+\frac{y_1}{\frac13}\cdot(-1)=1$$

Simplify and we get:

$$2x_1-3y_1=1\tag{2}$$

To find the find its coordinate $(x_1, y_1)$ for point $B$, we need another equation. Note that point $B$ is on the ellipse, hence

$$2x_1^2 + 9y_1^2 = 3\tag{3}$$

Combining (2) and (3), we get

$$(x_1, y_1)=(-\frac13, -\frac59)~~~~\text{or}~~~~(x_1, y_1)=(1, \frac13)$$

Namely, we get two tangent points on the ellipse,

$$B_1(-\frac13, -\frac59)~~~~\text{or}~~~~B_2(1, \frac13)$$

Finally, let's find these two tangent lines.

From point $A(3, -1)$ and point $B_1(-\frac13, -\frac59)$, we get tangent line $l_{AB_1}$

$$\boxed{y=-\frac2{15}x-\frac35}$$

From point $A(3, -1)$ and point $B_2(1, \frac13)$, we get tangent line $l_{AB_2}$

$$\boxed{y=-\frac23x+1}$$

Answer 2

It seems you've been exposed to some part of Joachimsthal's notation, in particular the $s_1=0,$ which is the tangent when the point is on the conic, but which more generally defines the polar of the point when the point is not on the conic.

There are more Joachimsthal notations. $s_1^2=s \cdot s_{11}$ is the one for your situation.

Let's factor $s_1^2-s \cdot s_{11},$ where the point is $(3,-1)$ and $s=0$ is your ellipse equation.

$$(2x^2+9y^2-3)(2(3)^2+9(-1)^2-3)-(2x\cdot 3+9y\cdot (-1)-3)^2)=\\3\cdot (3y+2x-3)(15y+2x+9)$$

As you can see, the two tangents fall right out.