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Example 1#

Given an ellipse $C:\frac{x^2}{4}+\frac{y^2}{3}=1$, if the line $l:y=kx+m$ intersects the ellipse $C$ at points $A, B$ (where $A,B$ are not the vertices), and a circle with diameter $AB$ passes through $(0,0)$. Prove that the line $l$ is tangent to the circle.

Analysis#

From the given information, we can see that $OA\bot OB$, which means $k_{OA}\bullet k_{OB}=-1$. Therefore, we can construct a quadratic equation in terms of $\frac{y}{x}$ and solve it using the Vieta's theorem.

Solution#

Rewrite the equation of the ellipse as

$3x^2+4y^2=12$

Since the line $l$ does not pass through the origin, we can assume the equation of the line $l$ as $mx+ny=1$. Therefore¹,

$3x^2+4y^2=12\bullet 1^2=12\bullet {(mx+ny)}^2$

Simplifying, we get

$(12n^2-4)y^2+24mn\bullet xy+(12m^2-3)x^2=0$

Dividing both sides of the equation by $x^2$, we get

$(12n^2-4)\frac{y^2}{x^2}+24mn\bullet\frac{y}{x}+(12m^2-3)=0$

Clearly, $\Delta>0$. Using Vieta's theorem, we have

$k_{OA}\bullet k_{OB}=\frac{12m^2-3}{12n^2-4}=-1$

Therefore,

$12(m^2+n^2)=7$

The distance from the origin $(0,0)$ to the line $l$ is

$d=\frac{1}{\sqrt{m^2+n^2}}=\sqrt{\frac{12}{7}}=\frac{\sqrt{84}}{7}$

Hence, the distance from the line $l$ to $(0,0)$ is constant, which means the line $l$ is tangent to the circle with center $(0,0)$ and radius $\frac{\sqrt{84}}{7}$.

The above example demonstrates the basic idea of homogenization in translation. But what is "translation"? And how is it used? Let's look at the following example.

Example 2#

Given an ellipse $C:\frac{x^2}{4}+\frac{y^2}{3}=1$, if the line $l:y=kx+b$ intersects the ellipse $C$ at points $A, B$ (where $A, B$ are not the vertices), and a circle with diameter $AB$ passes through the right vertex of the ellipse. Prove that the line $l$ passes through a fixed point, and find the coordinates of that point.

Analysis#

By drawing a diagram, we can see that the "origin $O$" in the previous example has become the right vertex of the ellipse (denoted as point $D$), and $k_{DA}\bullet k_{DB}$ is still equal to the constant $-1$. Therefore, we can consider transforming point $D$ into the origin, i.e., "translating the coordinate system".

Solution#

Translate the coordinate system two units to the right, as shown in the diagram. In the new coordinate system, the equation of the ellipse becomes

$\frac{{(x + 2)}^2}{4} + \frac{y^2}{3}=1$

That is,

$3x^2+12x+4y^2=0$

Since the line $l$ does not pass through the origin, we can assume the equation of the line $l$ as $mx+ny=1$. Therefore,

$3x^2+12x\bullet(mx+ny)+4y^2=0$

Simplifying, we get

$4y^2+12n\bullet xy+(12m+3)x^2=0$

Dividing both sides of the equation by $x^2$, we get

$4\frac{y^2}{x^2}+12n\bullet\frac{y}{x}+(12m+3)=0$

Since the circle with diameter $AB$ passes through the right vertex of the ellipse, we have

$k_{DA}\bullet k_{DB}=-1$

Clearly, $\Delta>0$. Using Vieta's theorem, we have

$k_{DA}\bullet k_{DB}=\frac{12m+3}{4}=-1$

Solving for $m$, we get

$m=-\frac{7}{12}$

Substituting this value into the equation of the line, we get

$-\frac{7}{12}x+ny=1$

Let $y=0$, then

$x=-\frac{12}{7}$

Therefore, the line $l$ passes through the fixed point $(-\frac{12}{7}, 0)$, which corresponds to the point $(\frac{2}{7}, 0)$ in the original coordinate system.

Through the above two examples, we have explained the basic principle of using translation homogenization. This method greatly reduces the difficulty of calculations. In translation homogenization, "translation" is used to achieve "homogenization". In fact, homogenization can also be constructed without translation.

Example 3#

Given hyperbola $C$ has the same asymptotes as hyperbola $\frac{x^2}{4}-\frac{y^2}{3}=1$, and point $A(2,3)$ lies on $C$. The line $l$ intersects hyperbola $C$ at points $P, Q$, and the lines $AP, AQ$ are symmetric with respect to the line $x=2$. Prove that the line $l$ passes through a fixed point.

Analysis#

Assume $A(x_1,y_1),B(x_2,y_2)$. Then $k_{AP}=\frac{y_1-3}{x_1},k_{AQ}=\frac{y_2-3}{x_2}$. To avoid translation, we need to construct a quadratic equation in terms of $\frac{y-3}{x}$. The line $l$ can be assumed as $mx+n(y-3)=1$, and the equation of the hyperbola can be simplified as $\frac{{(x + 2)}^ 2 }{ 4 } + \frac{y ^2}{3}=1$. By treating $y-3$ as a whole, we have $\frac{{(x + 2)}^ 2 }{ 4 } + \frac{ (y - 3 ) ^ 2 }{ 3 } =1$. Homogenization can be used to solve this equation.

Solution#

Rewrite the equation of the hyperbola as

$\frac{{(x + 2)}^ 2 }{ 4 } + \frac{ (y - 3 ) ^ 2 }{ 3 } =1$

Since the line $l$ does not pass through the origin, we can assume the equation of the line $l$ as $mx+n(y-3)=1$. Therefore,

$\frac{{(x + 2)}^ 2 }{ 4 } + \frac{ (y - 3 ) ^ 2 }{ 3 } \bullet (mx+n(y-3))=1$

Simplifying, we get

$(4n-1)(y-3)^2+4(mx+n(y-3))x+\frac{x^2}{4}=0$

Dividing both sides of the equation by $x^2$, we get

$(4n-1)\frac{(y-3)^2}{x^2}+4(mx+n(y-3))\frac{y-3}{x}+\frac{1}{4}=0$

Clearly, $\Delta>0$. Using Vieta's theorem, we have

$k_{AP}+k_{AQ}=-\frac{4(mx+n(y-3))}{4n-1}=-1$

Therefore,

$4(mx+n(y-3))=4n-1$

Substituting the equation of the line $l$, we get

$4mx+4ny-12n=4n-1$

Simplifying, we get

$4mx+4ny=1$

Dividing both sides of the equation by $4$, we get

$mx+ny=\frac{1}{4}$

This equation represents a line passing through a fixed point.

Thus, we have shown that the line $l$ passes through a fixed point.