Question Statement

Find the points of intersection of the line $x-2y+3=0$ and the parabola $y^2=8x+1$. Also, find the length of the chord intercepted by the parabola on this line and determine if this chord is a focal chord.

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Background and Explanation

To find the intersection points of a line and a parabola, we solve their equations as a system. The distance between these intersection points represents the length of the chord. A chord is considered a "focal chord" if it passes through the focus of the parabola.

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Solution

Part (a): Finding Points of Intersection

To find where the line and parabola meet, we solve the equations simultaneously.

Given the line: $x-2y+3=0$ And the parabola: $y^2=8x+1$

From equation (i), we can express $x$ in terms of $y$: $x=2y-3$

Now, substitute this expression for $x$ into equation (ii): $y^2=8(2y-3)+1$ $y^2=16y-24+1$ $y^2-16y+23=0$

We solve this quadratic equation using the quadratic formula $y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$: $y=\frac{-(-16)\pm \sqrt{(-16{)}^2-4(1)(23)}}{2(1)}$ $y=\frac{16\pm \sqrt{256-92}}{2}$ $y=\frac{16\pm \sqrt{164}}{2}$ $y=\frac{16\pm 2\sqrt{41}}{2}$ $y=8\pm \sqrt{41}$

So, the $y$-coordinates are $y_1=8-\sqrt{41}$ and $y_2=8+\sqrt{41}$.

Next, we find the corresponding $x$-coordinates using $x=2y-3$:

For $y=8-\sqrt{41}$: $x=2(8-\sqrt{41})-3=16-2\sqrt{41}-3=13-2\sqrt{41}$

For $y=8+\sqrt{41}$: $x=2(8+\sqrt{41})-3=16+2\sqrt{41}-3=13+2\sqrt{41}$

The points of intersection are: $A(13-2\sqrt{41},8-\sqrt{41})\text{and}B(13+2\sqrt{41},8+\sqrt{41})$

Part (b): Finding the Length of the Chord

The length of the chord is the distance between points $A$ and $B$. We use the distance formula: $|AB|=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

Substitute the coordinates: $|AB|=\sqrt{[(13+2\sqrt{41})-(13-2\sqrt{41}){]}^2+[(8+\sqrt{41})-(8-\sqrt{41}){]}^2}$ $|AB|=\sqrt{(4\sqrt{41}{)}^2+(2\sqrt{41}{)}^2}$ $|AB|=\sqrt{16(41)+4(41)}$ $|AB|=\sqrt{656+164}$ $|AB|=\sqrt{820}$ $|AB|=2\sqrt{205}$

Part (c): Determining if the Chord is a Focal Chord

A chord is a focal chord if the line passes through the focus of the parabola. First, we find the focus of $y^2=8x+1$.

Rewrite the parabola in standard form $Y^2=4aX$: $y^2=8\left(x+\frac{1}{8}\right)$

Let $Y=y$ and $X=x+\frac{1}{8}$. Comparing $Y^2=8X$ with $Y^2=4aX$: $4a=8⟹a=2$

The focus of the standard parabola $Y^2=4aX$ is $(X,Y)=(a,0)$. Setting $X=2$: $x+\frac{1}{8}=2⟹x=2-\frac{1}{8}=\frac{15}{8}$ Setting $Y=0$: $y=0$

The focus is $\left(\frac{15}{8},0\right)$. Now, we check if this point lies on the line $x-2y+3=0$: $\frac{15}{8}-2(0)+3=\frac{15}{8}+\frac{24}{8}=\frac{39}{8}$

Since $\frac{39}{8}\ne 0$, the focus does not lie on the line. Therefore, the chord is not a focal chord.

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Key Formulas or Methods Used

• Substitution Method: For solving systems of equations.
• Quadratic Formula: $y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
• Distance Formula: $d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$
• Parabola Focus: For $y^2=4ax$, the focus is $(a,0)$. For shifted parabolas, use $X=a$ and $Y=0$.

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Summary of Steps

1. Isolate $x$ in the linear equation and substitute it into the parabola equation.
2. Solve the resulting quadratic equation to find the $y$-coordinates of the intersection points.
3. Calculate the corresponding $x$-coordinates to identify points $A$ and $B$.
4. Apply the distance formula to find the length of the segment $AB$.
5. Determine the focus of the parabola by converting the equation to standard form.
6. Test if the focus coordinates satisfy the original line equation to check if the chord is a focal chord.