Understanding Key Concepts and Formulas for Ellipses

To solve this problem, we need a solid understanding of the properties of an ellipse and the formula for its normal.

An ellipse with its center at the origin $(0,0)$ has two standard forms, depending on whether its major axis lies along the x-axis or the y-axis.

1. Standard Equation of an Ellipse:

   • If the major axis is along the x-axis (meaning $a > b$, where $a$ is the semi-major axis and $b$ is the semi-minor axis), the equation is: ${x^2 \over a^2} + {y^2 \over b^2} = 1$
   • If the major axis is along the y-axis (meaning $a > b$), the equation is: ${x^2 \over b^2} + {y^2 \over a^2} = 1$

2. Eccentricity ($e$): This value quantifies the "roundness" of the ellipse. For an ellipse, $0 < e < 1$.

3. Relationship between $a$, $b$, and $e$: These fundamental parameters are related by the equation: $b^2 = a^2(1 - e^2)$ This formula is valid regardless of the orientation of the major axis, provided $a$ is always the semi-major axis.

4. Directrices: An ellipse has two directrices, which are lines perpendicular to the major axis.

   • If the major axis is along the x-axis, the directrices are vertical lines given by $x = \pm {a \over e}$.
   • If the major axis is along the y-axis, the directrices are horizontal lines given by $y = \pm {a \over e}$.

5. Equation of the Normal to an Ellipse at a Point P($x_1, y_1$): For an ellipse ${x^2 \over a^2} + {y^2 \over b^2} = 1$ (where $a^2$ is under $x^2$ and $b^2$ under $y^2$), the equation of the normal at a point $P(x_1, y_1)$ on the ellipse is given by: ${a^2x \over x_1} - {b^2y \over y_1} = a^2 - b^2$ This formula is crucial for the final step of our problem.

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Step-by-Step Solution Strategy

Our approach to solving this problem will involve three distinct phases:

1. Determine the specific equation of the ellipse: We'll use the given information about its center, directrix, and eccentricity.
2. Find the complete coordinates of point P: Since $P(1, \beta)$ lies on the ellipse, its coordinates must satisfy the ellipse's equation.
3. Calculate the equation of the normal: With the ellipse's equation and the exact coordinates of P, we can apply the normal formula.

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Step 1: Determine the Equation of the Ellipse

Let's begin by extracting the given information and using it to define our ellipse.

• Given Information:

  • Center of the ellipse: Origin $(0,0)$.
  • Eccentricity: $e = {1 \over 2}$.
  • One directrix: $x = 4$.

• Identify the Orientation of the Major Axis: The directrix is given as $x=4$. This is a vertical line. For an ellipse centered at the origin, vertical directrices ($x = \pm k$) indicate that the major axis of the ellipse lies along the x-axis. If the major axis were along the y-axis, the directrices would be horizontal lines ($y = \pm k$). Therefore, the standard form of our ellipse is: ${x^2 \over a^2} + {y^2 \over b^2} = 1$ where $a$ is the semi-major axis and $b$ is the semi-minor axis.

• Use the Directrix Equation to find the semi-major axis 'a': For an ellipse with its major axis along the x-axis, the equations of the directrices are $x = \pm {a \over e}$. We are given one directrix as $x=4$. So, we can write: $4 = {a \over e}$ Now, substitute the given eccentricity $e = {1 \over 2}$: $4 = {a \over {1/2}}$ $4 = 2a$ Solving for $a$: $a = 2$ This means the length of the semi-major axis is 2 units. Consequently, $a^2 = (2)^2 = 4$.

• Calculate $b^2$ using the fundamental relationship $b^2 = a^2(1 - e^2)$: We have $a=2$ (so $a^2=4$) and $e={1 \over 2}$. We can now find $b^2$: $b^2 = a^2(1 - e^2)$ Substitute the values: $b^2 = 4 \left(1 - \left({1 \over 2}\right)^2\right)$ $b^2 = 4 \left(1 - {1 \over 4}\right)$ $b^2 = 4 \left({4-1 \over 4}\right)$ $b^2 = 4 \left({3 \over 4}\right)$ $b^2 = 3$

• Write the Equation of the Ellipse: Now that we have $a^2 = 4$ and $b^2 = 3$, we can substitute these values into the standard equation ${x^2 \over a^2} + {y^2 \over b^2} = 1$: ${x^2 \over 4} + {y^2 \over 3} = 1$ This is the specific equation of the ellipse we are working with.

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Step 2: Find the Coordinates of Point P

We are given that point $P(1, \beta)$ lies on this ellipse, and we know that $\beta > 0$. If a point lies on an ellipse, its coordinates must satisfy the ellipse's equation.

• Substitute P's coordinates into the ellipse equation: Substitute $x=1$ and $y=\beta$ into the ellipse equation ${x^2 \over 4} + {y^2 \over 3} = 1$: ${(1)^2 \over 4} + {(\beta)^2 \over 3} = 1$ ${1 \over 4} + {{\beta^2} \over 3} = 1$

• Solve for $\beta$: First, isolate the term containing $\beta^2$: ${{\beta^2} \over 3} = 1 - {1 \over 4}$ ${{\beta^2} \over 3} = {4 \over 4} - {1 \over 4}$ ${{\beta^2} \over 3} = {3 \over 4}$ Now, multiply both sides by 3 to find $\beta^2$: $\beta^2 = 3 \times {3 \over 4}$ $\beta^2 = {9 \over 4}$ Take the square root of both sides to find $\beta$: $\beta = \pm \sqrt{{9 \over 4}}$ $\beta = \pm {3 \over 2}$ The problem statement specifies that $\beta > 0$. Therefore, we choose the positive value: $\beta = {3 \over 2}$ So, the exact coordinates of point P are $P\left(1, {3 \over 2}\right)$. This will be our $(x_1, y_1)$ for the normal equation.

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Step 3: Determine the Equation of the Normal at P

Now that we have the ellipse's parameters ($a^2, b^2$) and the exact coordinates of point P ($x_1, y_1$), we can apply the formula for the equation of the normal.

• Recall the Normal Equation Formula: For the ellipse ${x^2 \over a^2} + {y^2 \over b^2} = 1$, the equation of the normal at a point $(x_1, y_1)$ is: ${a^2x \over x_1} - {b^2y \over y_1} = a^2 - b^2$

• Identify the Necessary Values: From our previous calculations:

  • $a^2 = 4$
  • $b^2 = 3$
  • $(x_1, y_1) = \left(1, {3 \over 2}\right)$

• Substitute Values and Simplify: Substitute these values into the normal equation formula: ${4x \over 1} - {3y \over {{3 \over 2}}} = 4 - 3$ Simplify the terms. Remember that dividing by a fraction is the same as multiplying by its reciprocal: $4x - \left(3y \times {2 \over 3}\right) = 1$ $4x - 2y = 1$

This is the equation of the normal to the ellipse at point P.

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Final Answer Verification and Common Pitfalls

The equation of the normal to the ellipse at point P is $4x - 2y = 1$.

Let's compare this with the given options: (A) $4x – 3y = 2$ (B) $8x – 2y = 5$ (C) $7x – 4y = 1$ (D) $4x – 2y = 1$

Our derived equation matches option (D).

Important Note on Discrepancy: The problem statement indicates that option (A) is the correct answer. However, based on a rigorous step-by-step derivation using standard formulas, the result $4x - 2y = 1$ (Option D) is consistently obtained. It is possible there is a typo in the provided "Correct Answer" for this question. Always trust your calculations if you are confident in the formulas and steps.

Common Pitfalls to Avoid:

• Incorrect Orientation: Misidentifying the major axis (e.g., assuming x-axis major axis when it's y-axis) can lead to wrong $a$ and $b$ values or incorrect directrix formulas. A directrix $x=k$ always implies the major axis is along the x-axis for an origin-centered ellipse.
• Mixing up $a^2$ and $b^2$: In the standard ellipse equation $x^2/A + y^2/B = 1$, $A$ is $a^2$ if $A>B$, and $B$ is $a^2$ if $B>A$. For an ellipse with major axis along x-axis, $a^2$ is under $x^2$.
• Formula for Normal vs. Tangent: Ensure you use the correct formula for the normal. The tangent equation at $(x_1, y_1)$ is $xx_1/a^2 + yy_1/b^2 = 1$.
• Algebraic Errors: Be careful with fractions and signs during substitution and simplification.

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Key Takeaways

This problem is a comprehensive test of your understanding of ellipses, requiring you to:

1. Interpret geometric properties: Correctly deduce the ellipse's orientation and parameters ($a, b$) from its directrix and eccentricity.
2. Apply fundamental relationships: Use $b^2 = a^2(1-e^2)$ to find the complete equation of the ellipse.
3. Utilize point-on-curve property: Substitute a given point's coordinates into the ellipse equation to find missing values.
4. Master coordinate geometry formulas: Accurately apply the formula for the equation of the normal to an ellipse at a given point.