Key Concept: Equation of a Chord with a Given Midpoint

In the study of conic sections, a powerful and frequently used formula to find the equation of a chord whose midpoint is given is $T = S_1$. This formula is applicable to all conic sections (parabola, ellipse, hyperbola, and circles).

Let's define the terms:

• $S=0$: This represents the general equation of the conic section. For a parabola like $x^2=8y$, we rewrite it as $S = x^2 - 8y = 0$.
• $(x_1, y_1)$: This is the given midpoint of the chord.
• $T$: This expression is derived from the equation $S$ by applying specific transformation rules. These rules essentially convert quadratic terms into linear terms, similar to how tangent equations are formed:
  • Replace $x^2$ with $xx_1$
  • Replace $y^2$ with $yy_1$
  • Replace $x$ with $\frac{x+x_1}{2}$
  • Replace $y$ with $\frac{y+y_1}{2}$
  • Replace $xy$ with $\frac{xy_1+yx_1}{2}$
  • Constant terms remain unchanged.
• $S_1$: This is the numerical value obtained by substituting the coordinates of the midpoint $(x_1, y_1)$ directly into the equation $S$. That is, if $S = Ax^2 + Bxy + Cy^2 + Dx + Ey + F$, then $S_1 = Ax_1^2 + Bx_1y_1 + Cy_1^2 + Dx_1 + Ey_1 + F$.

The formula $T=S_1$ works because it inherently captures the condition that the midpoint $(x_1, y_1)$ equally divides the segment joining the two points where the chord intersects the conic. It provides a direct path to the chord's equation without needing to calculate the endpoints of the chord.

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Step-by-Step Solution

Step 1: Identify the Parabola and Midpoint for the Chord Calculation

The problem states that the point $P(\alpha, \beta)$ lies on a chord of the parabola $x^2=8y$. The midpoint of this chord is given as $\left(1, \frac{5}{4}\right)$. Our first goal is to find the equation of this chord.

• Equation of the Parabola ($S=0$): We rewrite $x^2 = 8y$ as $S = x^2 - 8y = 0$.
• Midpoint of the Chord ($(x_1, y_1)$): We are given $(x_1, y_1) = \left(1, \frac{5}{4}\right)$.

Step 2: Apply the Chord Formula $T = S_1$

We will now calculate $T$ and $S_1$ using the parabola $S = x^2 - 8y$ and the midpoint $(x_1, y_1) = \left(1, \frac{5}{4}\right)$.

• Calculate $T$: We apply the transformation rules to $S = x^2 - 8y$:

  • The term $x^2$ transforms to $xx_1$.
  • The term $-8y$ transforms to $-8\left(\frac{y+y_1}{2}\right)$.

  Substituting $x_1=1$ and $y_1=\frac{5}{4}$: $T = x(1) - 8\left(\frac{y + \frac{5}{4}}{2}\right)$ $T = x - 4\left(y + \frac{5}{4}\right)$ $T = x - 4y - 4\left(\frac{5}{4}\right)$ $T = x - 4y - 5$

• Calculate $S_1$: We substitute the midpoint coordinates $(x_1, y_1) = \left(1, \frac{5}{4}\right)$ into the parabola equation $S = x^2 - 8y$: $S_1 = (x_1)^2 - 8(y_1)$ $S_1 = (1)^2 - 8\left(\frac{5}{4}\right)$ $S_1 = 1 - (2 \times 5)$ $S_1 = 1 - 10$ $S_1 = -9$

• Form the Equation of the Chord: Now, we equate $T = S_1$: $x - 4y - 5 = -9$ Rearranging the terms to get the standard linear equation of the chord: $x - 4y + 4 = 0 \quad \cdots (i)$ This is the line on which point $P(\alpha, \beta)$ lies.

Step 3: Establish Equations for Point $P(\alpha, \beta)$

The problem states that point $P(\alpha, \beta)$ satisfies two conditions:

1. It lies on the chord whose equation we just found.
2. It lies on the parabola $y^2 = 4x$.

• Condition 1: $P(\alpha, \beta)$ lies on the chord (i) Substituting $(\alpha, \beta)$ into the chord equation $x - 4y + 4 = 0$: $\alpha - 4\beta + 4 = 0 \quad \cdots (ii)$

• Condition 2: $P(\alpha, \beta)$ lies on the parabola $y^2 = 4x$ Substituting $(\alpha, \beta)$ into the parabola equation $y^2 = 4x$: $\beta^2 = 4\alpha \quad \cdots (iii)$

Step 4: Solve the System of Equations for $\alpha$ and $\beta$

We now have a system of two equations with two variables:

1. $\alpha - 4\beta + 4 = 0$
2. $\beta^2 = 4\alpha$

We will use substitution to solve for $\alpha$ and $\beta$. From equation (ii), it is straightforward to express $\alpha$ in terms of $\beta$: $\alpha = 4\beta - 4$

Now, substitute this expression for $\alpha$ into equation (iii): $\beta^2 = 4(4\beta - 4)$ Expand and rearrange the terms to form a standard quadratic equation in $\beta$: $\beta^2 = 16\beta - 16$ $\beta^2 - 16\beta + 16 = 0$

We solve this quadratic equation using the quadratic formula, $\beta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a=1$, $b=-16$, $c=16$: $\beta = \frac{-(-16) \pm \sqrt{(-16)^2 - 4(1)(16)}}{2(1)}$ $\beta = \frac{16 \pm \sqrt{256 - 64}}{2}$ $\beta = \frac{16 \pm \sqrt{192}}{2}$

To simplify $\sqrt{192}$, we look for the largest perfect square factor of 192. We know $192 = 64 \times 3$. So, $\sqrt{192} = \sqrt{64 \times 3} = \sqrt{64} \times \sqrt{3} = 8\sqrt{3}$.

Substitute this back into the expression for $\beta$: $\beta = \frac{16 \pm 8\sqrt{3}}{2}$ $\beta = 8 \pm 4\sqrt{3}$

This gives us two possible values for $\beta$:

• $\beta_1 = 8 + 4\sqrt{3}$
• $\beta_2 = 8 - 4\sqrt{3}$

Now, we find the corresponding values of $\alpha$ using the relation $\alpha = 4\beta - 4$:

• For $\beta_1 = 8 + 4\sqrt{3}$: $\alpha_1 = 4(8 + 4\sqrt{3}) - 4$ $\alpha_1 = 32 + 16\sqrt{3} - 4$ $\alpha_1 = 28 + 16\sqrt{3}$ So, one possible point is $P_1 = (28 + 16\sqrt{3}, 8 + 4\sqrt{3})$.

• For $\beta_2 = 8 - 4\sqrt{3}$: $\alpha_2 = 4(8 - 4\sqrt{3}) - 4$ $\alpha_2 = 32 - 16\sqrt{3} - 4$ $\alpha_2 = 28 - 16\sqrt{3}$ So, the other possible point is $P_2 = (28 - 16\sqrt{3}, 8 - 4\sqrt{3})$.

Step 5: Calculate the Required Expression $(\alpha-28)(\beta-8)$

The problem asks for the value of the expression $(\alpha-28)(\beta-8)$. Let's calculate this for both points we found.

• For $P_1 = (28 + 16\sqrt{3}, 8 + 4\sqrt{3})$: First, determine the terms $(\alpha-28)$ and $(\beta-8)$: $\alpha - 28 = (28 + 16\sqrt{3}) - 28 = 16\sqrt{3}$ $\beta - 8 = (8 + 4\sqrt{3}) - 8 = 4\sqrt{3}$ Now, multiply these terms: $(\alpha-28)(\beta-8) = (16\sqrt{3})(4\sqrt{3})$ $= (16 \times 4) \times (\sqrt{3} \times \sqrt{3})$ $= 64 \times 3$ $= 192$

• For $P_2 = (28 - 16\sqrt{3}, 8 - 4\sqrt{3})$: First, determine the terms $(\alpha-28)$ and $(\beta-8)$: $\alpha - 28 = (28 - 16\sqrt{3}) - 28 = -16\sqrt{3}$ $\beta - 8 = (8 - 4\sqrt{3}) - 8 = -4\sqrt{3}$ Now, multiply these terms: $(\alpha-28)(\beta-8) = (-16\sqrt{3})(-4\sqrt{3})$ $= (-16 \times -4) \times (\sqrt{3} \times \sqrt{3})$ $= 64 \times 3$ $= 192$

Both possible points for $P(\alpha, \beta)$ yield the same value for the expression $(\alpha-28)(\beta-8)$. This consistency confirms our calculations.

The final answer is $\boxed{192}$.

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Tips and Common Mistakes to Avoid

• Distinguish $T=S_1$ from $T=0$: A common pitfall is confusing $T=S_1$ with $T=0$.
  • $T=0$ is the equation of the tangent at $(x_1, y_1)$ if $(x_1, y_1)$ lies on the conic. If $(x_1, y_1)$ is outside the conic, $T=0$ is the chord of contact.
  • $T=S_1$ is specifically used when $(x_1, y_1)$ is the midpoint of a chord.
• Accurate Transformation Rules for $T$: Be very careful when applying the rules for $T$. The most common error is forgetting the division by 2 for the linear terms ($x \to \frac{x+x_1}{2}$ and $y \to \frac{y+y_1}{2}$). Forgetting this will lead to an incorrect chord equation.
• Algebraic Precision: Double-check all substitutions and simplifications, especially when dealing with fractions and square roots. A small arithmetic error can propagate through the entire solution.
• Understanding the Question: The problem involves two different parabolas. Make sure you correctly identify which parabola belongs to the chord with the given midpoint ($x^2=8y$) and which parabola the point $P$ itself lies on ($y^2=4x$).

Summary and Key Takeaway

This problem effectively tests the understanding and application of the $T=S_1$ formula for finding the equation of a chord with a given midpoint. Once the chord equation is established, the problem reduces to solving a system of equations formed by the chord and the second parabola to find the coordinates of point $P$. Finally, careful evaluation of the desired expression using the calculated coordinates leads to the answer. The ability to correctly apply conic section formulas and execute algebraic manipulations is crucial for solving such problems.