Key Concepts and Formulas

This problem requires a strong understanding of the standard equation of an ellipse and, critically, the equation of the normal line to an ellipse at a given point. We will also use the fundamental concept that if a point lies on a curve or line, its coordinates must satisfy the equation of that curve or line.

1. Standard Equation of an Ellipse: An ellipse centered at the origin with its axes along the coordinate axes has the standard form: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ Here, $a^2$ and $b^2$ represent the squares of the lengths of the semi-axes. To use the normal equation formula correctly, it's crucial to identify $a^2$ and $b^2$ from the given ellipse equation. For example, if the equation is $Ax^2 + By^2 = C$, we divide by $C$ to get $\frac{A}{C}x^2 + \frac{B}{C}y^2 = 1$, which can be written as $\frac{x^2}{C/A} + \frac{y^2}{C/B} = 1$. Thus, $a^2 = C/A$ and $b^2 = C/B$.

2. Equation of the Normal to an Ellipse: The equation of the normal line to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at a point $P(x_1, y_1)$ lying on the ellipse is given by: $\frac{a^2x}{x_1} - \frac{b^2y}{y_1} = a^2 - b^2$ This formula is derived from finding the slope of the tangent at $(x_1, y_1)$ using calculus (differentiation) and then using the negative reciprocal for the normal's slope. It's a fundamental formula in conic sections.

3. Intercepts of a Line:

   • To find the x-intercept of a line, set $y=0$ in its equation and solve for $x$.
   • To find the y-intercept of a line, set $x=0$ in its equation and solve for $y$.

Step-by-Step Solution

Let's systematically solve the problem by breaking it down into logical steps to find the value of $\beta$.

Step 1: Standardize the Ellipse Equation and Identify Parameters

The given equation of the ellipse is $2x^2 + y^2 = 1$. Why this step is necessary: To effectively use the standard formula for the normal equation, we must first express the given ellipse equation in its standard form, $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. This allows us to correctly identify the values of $a^2$ and $b^2$.

We can rewrite $2x^2$ as $\frac{x^2}{1/2}$ and $y^2$ as $\frac{y^2}{1}$. So, the ellipse equation becomes: $\frac{x^2}{1/2} + \frac{y^2}{1} = 1$ By comparing this with the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, we can identify the parameters: $a^2 = \frac{1}{2}$ $b^2 = 1$ (Note: Since $b^2 > a^2$, the major axis of this ellipse is along the y-axis, but this fact is not directly used in finding $\beta$.)

Step 2: Formulate the General Equation of the Normal to the Ellipse

Let $P(x_1, y_1)$ be the point of intersection of the line $y=mx$ and the ellipse in the first quadrant. This information is crucial because it implies that $x_1 > 0$ and $y_1 > 0$. Why this step is necessary: We need an equation that describes the normal line at point $P(x_1, y_1)$ so we can use the given intercept points to find $x_1$, $y_1$, and eventually $\beta$.

Now, we use the formula for the equation of the normal to the ellipse at $P(x_1, y_1)$: $\frac{a^2x}{x_1} - \frac{b^2y}{y_1} = a^2 - b^2$ Substitute the identified values of $a^2 = \frac{1}{2}$ and $b^2 = 1$ into this formula: $\frac{(1/2)x}{x_1} - \frac{(1)y}{y_1} = \frac{1}{2} - 1$ $\frac{x}{2x_1} - \frac{y}{y_1} = -\frac{1}{2}$ This is the specific equation of the normal line to our given ellipse at any point $P(x_1, y_1)$ on it. Let's call this Equation (1).

Step 3: Use the X-intercept of the Normal to Determine $x_1$

We are given that the normal to the ellipse at P meets the x-axis at $\left( { - {1 \over {3\sqrt 2 }},0} \right)$. Why this step is necessary: Since this point lies on the normal line, its coordinates must satisfy the normal's equation (Equation 1). By substituting the x-intercept's coordinates, we can solve for the unknown $x_1$.

Substitute $x = -{1 \over {3\sqrt 2 }}$ and $y = 0$ into Equation (1): $\frac{{ - {1 \over {3\sqrt 2 }}}}{2x_1} - \frac{0}{y_1} = -\frac{1}{2}$ $\frac{{ - 1}}{{6\sqrt 2 x_1}} = -\frac{1}{2}$ Now, we solve for $x_1$: Multiply both sides by $-1$ to simplify: $\frac{1}{{6\sqrt 2 x_1}} = \frac{1}{2}$ Cross-multiply to isolate $x_1$: $2 = 6\sqrt 2 x_1$ $x_1 = \frac{2}{6\sqrt 2}$ $x_1 = \frac{1}{3\sqrt 2}$ Since point P is in the first quadrant, its x-coordinate $x_1$ must be positive, which our calculated value $\frac{1}{3\sqrt 2}$ confirms.

Step 4: Use the Ellipse Equation to Determine $y_1$

The point $P(x_1, y_1)$ lies on the ellipse $2x^2 + y^2 = 1$. We have already found $x_1 = \frac{1}{3\sqrt 2}$. Why this step is necessary: Since P lies on the ellipse, its coordinates must satisfy the ellipse's equation. Having found $x_1$, we can now substitute it into the ellipse equation to find the corresponding $y_1$.

Substitute this value of $x_1$ into the ellipse equation: $2\left(x_1\right)^2 + \left(y_1\right)^2 = 1$ $2\left(\frac{1}{3\sqrt 2}\right)^2 + y_1^2 = 1$ First, calculate the square of $x_1$: $2\left(\frac{1^2}{(3\sqrt 2)^2}\right) + y_1^2 = 1$ $2\left(\frac{1}{9 \times 2}\right) + y_1^2 = 1$ $2\left(\frac{1}{18}\right) + y_1^2 = 1$ $\frac{1}{9} + y_1^2 = 1$ Now, solve for $y_1^2$: $y_1^2 = 1 - \frac{1}{9}$ $y_1^2 = \frac{9-1}{9}$ $y_1^2 = \frac{8}{9}$ Take the square root to find $y_1$: $y_1 = \sqrt{\frac{8}{9}}$ $y_1 = \frac{\sqrt{8}}{\sqrt{9}}$ $y_1 = \frac{2\sqrt{2}}{3}$ Since P is in the first quadrant, its y-coordinate $y_1$ must be positive, which our result $\frac{2\sqrt 2}{3}$ confirms. So, the point P is $\left( \frac{1}{3\sqrt 2}, \frac{2\sqrt 2}{3} \right)$.

Step 5: Use the Y-intercept of the Normal to Solve for $\beta$

We are given that the normal meets the y-axis at $(0, \beta)$. Why this step is necessary: This point also lies on the normal line. By substituting its coordinates ($x=0, y=\beta$) and the value of $y_1$ we just found into Equation (1), we can finally solve for $\beta$.

Substitute $x = 0$, $y = \beta$, and $y_1 = \frac{2\sqrt 2}{3}$ into Equation (1): $\frac{0}{2x_1} - \frac{\beta}{y_1} = -\frac{1}{2}$ $0 - \frac{\beta}{\frac{2\sqrt 2}{3}} = -\frac{1}{2}$ Simplify the complex fraction by multiplying $\beta$ by the reciprocal of $\frac{2\sqrt 2}{3}$: $-\frac{3\beta}{2\sqrt 2} = -\frac{1}{2}$ Multiply both sides by $-1$: $\frac{3\beta}{2\sqrt 2} = \frac{1}{2}$ Now, solve for $\beta$: Multiply both sides by $2\sqrt{2}$: $3\beta = \frac{2\sqrt 2}{2}$ $3\beta = \sqrt 2$ $\beta = \frac{\sqrt 2}{3}$

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Tips and Common Mistakes

• Standard Form is Crucial: Always begin by converting the ellipse equation to its standard form, $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. A common error is to mistake the coefficients of $x^2$ and $y^2$ directly as $a^2$ and $b^2$ without dividing by the constant term on the RHS. For $2x^2+y^2=1$, $a^2$ is $1/2$, not $2$.
• Normal Equation Formula: Ensure you accurately recall or derive the equation of the normal. Mistakes often occur with the signs or the order of $a^2$ and $b^2$ in the $a^2-b^2$ term.
• Quadrant Information: The problem states that point P is in the first quadrant. This implies $x_1 > 0$ and $y_1 > 0$. Use this information when taking square roots to select the correct positive values for $x_1$ and $y_1$.
• Algebraic Precision: Be meticulous with your calculations, especially when dealing with