Question Statement

Find the points of intersection of the line and the ellipse, and also find the length of the chord intercepted: $x-y+1=0$ $x^2+2y^2+3x-7y-11=0$

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Background and Explanation

To find the intersection points between a line and a curve (like an ellipse), we use the method of substitution. By expressing one variable in terms of the other from the linear equation and substituting it into the quadratic equation, we can solve for the coordinates. The distance formula is then applied to these points to determine the length of the chord.

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Solution

Given the equation of the line: $x-y+1=0\dots \text{(i)}$

And the equation of the ellipse: $x^2+2y^2+3x-7y-11=0\dots \text{(ii)}$

To find the points of intersection, we solve both equations simultaneously.

From equation (i), we can express $x$ in terms of $y$: $x=y-1$

Substitute this expression for $x$ into equation (ii): $(y-1{)}^2+2y^2+3(y-1)-7y-11=0$

Now, expand the terms and simplify: $(y^2-2y+1)+2y^2+(3y-3)-7y-11=0$ $y^2+2y^2-2y+3y-7y+1-3-11=0$ $3y^2-6y-13=0$

This is a quadratic equation in the form $a{y}^2+by+c=0$. We solve for $y$ using the quadratic formula: $y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ $y=\frac{-(-6)\pm \sqrt{(-6{)}^2-4(3)(-13)}}{2(3)}$ $y=\frac{6\pm \sqrt{36+156}}{6}$ $y=\frac{6\pm \sqrt{192}}{6}$

Since $\sqrt{192}=\sqrt{64\times 3}=8\sqrt{3}$: $y=\frac{6\pm 8\sqrt{3}}{6}$ $y=\frac{3\pm 4\sqrt{3}}{3}$

Finding the $x$-coordinates

Case 1: When $y=\frac{3-4\sqrt{3}}{3}$ Substitute $y$ into equation (i): $x-\left(\frac{3-4\sqrt{3}}{3}\right)+1=0$ $x=\frac{3-4\sqrt{3}}{3}-1$ $x=\frac{3-4\sqrt{3}-3}{3}$ $x=\frac{-4\sqrt{3}}{3}$

Case 2: When $y=\frac{3+4\sqrt{3}}{3}$ Substitute $y$ into equation (i): $x-\left(\frac{3+4\sqrt{3}}{3}\right)+1=0$ $x=\frac{3+4\sqrt{3}}{3}-1$ $x=\frac{3+4\sqrt{3}-3}{3}$ $x=\frac{4\sqrt{3}}{3}$

Thus, the points of intersection are: $A\left(\frac{-4\sqrt{3}}{3},\frac{3-4\sqrt{3}}{3}\right)\text{and}B\left(\frac{4\sqrt{3}}{3},\frac{3+4\sqrt{3}}{3}\right)$

Finding the Length of the Chord

The length of the chord is the distance between points $A$ and $B$: $|AB|=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$ $|AB|=\sqrt{{\left(\frac{4\sqrt{3}}{3}-\left(\frac{-4\sqrt{3}}{3}\right)\right)}^2+{\left(\frac{3+4\sqrt{3}}{3}-\frac{3-4\sqrt{3}}{3}\right)}^2}$ $|AB|=\sqrt{{\left(\frac{8\sqrt{3}}{3}\right)}^2+{\left(\frac{8\sqrt{3}}{3}\right)}^2}$ $|AB|=\sqrt{\frac{192}{9}+\frac{192}{9}}$ $|AB|=\sqrt{\frac{384}{9}}$ $|AB|=\sqrt{\frac{128}{3}}$

Simplifying the radical: $|AB|=\sqrt{\frac{64\times 2}{3}}=8\sqrt{\frac{2}{3}}$

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Key Formulas or Methods Used

• Substitution Method: Solving a system of equations by substituting one into another.
• Quadratic Formula: $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
• Distance Formula: $d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$
• Radical Simplification: Simplifying $\sqrt{192}$ and $\sqrt{384}$.

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Summary of Steps

1. Rearrange the linear equation to isolate $x$ ($x=y-1$).
2. Substitute this expression into the ellipse equation to create a quadratic equation in terms of $y$.
3. Simplify the quadratic equation to $3y^2-6y-13=0$.
4. Use the quadratic formula to find the two possible values for $y$.
5. Plug the $y$ values back into the linear equation to find the corresponding $x$ values.
6. Apply the distance formula to the two resulting points to calculate the chord length.