Let P(α, β) be a point on the parabola y² = 4x. If P also lies on the chord of the parabola x² = 8y

Q: Wait, why does the formula $T = S_1$ work for any conic section? Is it just a shortcut?

It is much more than a shortcut; it is a beautiful consequence of the definition of a chord. For a general conic $S=0$, the chord with midpoint $(x_1, y_1)$ is defined by the locus of points where the tangent-like expression $T$ equals the value of the conic at the midpoint $S_1$. It is a fundamental tool in coordinate geometry that bypasses the tedious process of finding the slope of the chord and the intersection points.

Q: Do I really need to find the exact values of $\alpha$ and $\beta$ to solve this?

In this specific problem, calculating the roots is quite straightforward, so finding them is a reliable path. However, notice that the final expression $(\alpha - 28)(\beta - 8)$ is symmetric. You could also use the properties of the roots of the quadratic equation (sum and product of roots) to evaluate the expression without explicitly calculating the roots, though calculating them here is very efficient.

Q: What if the point $P$ was not on the chord? Would the problem still be solvable?

If $P$ were not on the chord, the problem would be ill-posed. The constraint that $P$ lies on the chord is the critical link that allows us to connect the two parabolas. Without that, we would have an infinite number of points on the parabola $y^2 = 4x$ that satisfy no specific relationship with the chord.

Question

Let $P(\alpha, \beta)$ be a point on the parabola $y^2 = 4x$. If $P$ also lies on the chord of the parabola $x^2 = 8y$ whose mid point is $(1, \frac{5}{4})$. Then $(\alpha - 28)(\beta - 8)$ is equal to ______.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a coordinate plane, looking at two parabolas. One, $y^2 = 4x$, is a classic, opening its arms to the right. The other, $x^2 = 8y$, is reaching upwards.

They seem independent, yet they are bound together by a single point $P(\alpha, \beta)$ and a mysterious chord. This is the essence of JEE Advanced coordinate geometry—finding the hidden connections between seemingly separate entities.

Our journey begins with the chord of the parabola $x^2 = 8y$. We are given its midpoint, $M(1, \frac{5}{4})$.

### The Master Equation

In the heat of an exam, you might be tempted to find the slope of the chord or the coordinates of its endpoints. Resist that urge! Instead, reach for the most elegant tool in your arsenal: the formula $T = S_1$.

This formula is the secret key to unlocking any chord when the midpoint is known. For our parabola $x^2 - 8y = 0$, the expression $T$ is defined as $xx_1 - 4(y + y_1)$, and $S_1$ is simply the curve equation evaluated at the midpoint.

Substituting our midpoint $M(1, \frac{5}{4})$, we get:

$$x(1) - 4\left(y + \frac{5}{4}ight) = 1^2 - 8\left(\frac{5}{4}ight)$$

Simplifying this, we find $x - 4y - 5 = 1 - 10$, which leads us to the beautiful, simple linear equation:

$$x - 4y + 4 = 0$$

### The Algebraic Bridge

Now that we have the equation of the chord, $x - 4y + 4 = 0$, we know that our point $P(\alpha, \beta)$ must satisfy this equation. This gives us a direct relationship:

$$\alpha = 4\beta - 4$$

But $P$ is not just any point; it also lives on the first parabola, $y^2 = 4x$. This means $\beta^2 = 4\alpha$.

We have a system of two equations, and the path forward is clear: substitution. By replacing $\alpha$ with $4\beta - 4$ in the second equation, we get:

$$\beta^2 = 4(4\beta - 4)$$

$$\beta^2 = 16\beta - 16$$

Rearranging this, we arrive at the quadratic equation:

$$\beta^2 - 16\beta + 16 = 0$$

Using the quadratic formula, we find:

$$\beta = \frac{16 \pm \sqrt{256 - 64}}{2} = \frac{16 \pm \sqrt{192}}{2} = 8 \pm 4\sqrt{3}$$

### Final Calculation

We have our values for $\beta$, and we can easily find the corresponding $\alpha$ values using $\alpha = 4\beta - 4$. If $\beta = 8 \pm 4\sqrt{3}$, then:

$$\alpha = 4(8 \pm 4\sqrt{3}) - 4 = 28 \pm 16\sqrt{3}$$

The question asks us to evaluate $(\alpha - 28)(\beta - 8)$. Look at our results:

$$\alpha - 28 = \pm 16\sqrt{3}$$

$$\beta - 8 = \pm 4\sqrt{3}$$

When we multiply these, the signs align perfectly. We get:

$$(\pm 16\sqrt{3})(\pm 4\sqrt{3}) = 16 	imes 4 	imes 3 = 192$$

The complexity of the radicals vanishes, leaving behind a clean, solid integer. The final answer is **192**.

Animated Solution for Mathematics - Conic Sections: Let $P (α, β)$ be a point on the parabola $y^{2} = 4 x$ . If $P$ also lies on the chord of the parabola $x^{2} = 8 y$ whose mid point is $(1, \frac{5}{4})$ . Then $(α - 28) (β - 8)$ is equal to ______.

Enter Numerical Value:

Visualized Solution

Visualizing the Geometry

Equation of Chord: $T = S_{1}$

Substituting the Midpoint

Simplifying the Chord Equation

Point $P$ Constraints

Substitution for $β$

Forming the Quadratic Equation

Solving for $β$

Finding $α$

Evaluating the Target Expression

Final Calculation

The Sigma Insight: Chord in terms of Midpoint

Analyzing the Setup

The Master Equation

The Algebraic Bridge

Final Calculation

Similar Questions

The locus of the mid points of the chords of the hyperbola $x^{2} - y^{2} = 4$ , which touch the parabola $y^{2} = 8 x$ , is :

If a chord, which is not a tangent, of the parabola $y^{2} = 16 x$ has the equation $2 x + y = p$ , and midpoint $(h, k)$ , then which of the following is(are) possible value(s) of $p, h$ and $k$ ?

Let $P Q$ be a chord of the parabola $y^{2} = 12 x$ and the midpoint of $P Q$ be at $(4, 1)$ . Then, which of the following point lies on the line passing through the points $P$ and $Q$ ?

If $α x + β y = 109$ is the equation of the chord of the ellipse $\frac{x ^{2}}{9} + \frac{y ^{2}}{4} = 1.$ whose mid point is $(\frac{5}{2}, \frac{1}{2})$ then $α + β$ is equal to:

If the midpoint of a chord of the ellipseis $\frac{x ^{2}}{9} + \frac{y ^{2}}{4} = 1$ is $(2, 4/3)$ and the length of the chord is $\frac{2 α}{3}$ then a is:

The length of the chord of the ellipse $\frac{x ^{2}}{4} + \frac{y ^{2}}{2} = 1$ whose mid-point is $(1, \frac{1}{2})$ is:

The equation of the chord, of the ellipse $\frac{x ^{2}}{25} + \frac{y ^{2}}{16} = 1$ whose mid-point is $(3, 1)$ is:

If two distinct chords, drawn from the point $(p, q)$ on the circle $x^{2} + y^{2} = p x + q y$ (where $pq \neq = 0$ ) are bisected by the x-axis, then

Enter Numerical Value:

Visualized Solution

Visualizing the Geometry

Equation of Chord: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T=S1​

Substituting the Midpoint

Simplifying the Chord Equation

Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P Constraints

Substitution for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β

Forming the Quadratic Equation

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Evaluating the Target Expression

Final Calculation

The Sigma Insight: Chord in terms of Midpoint

Analyzing the Setup

The Master Equation

The Algebraic Bridge

Final Calculation

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Equation of Chord: $T = S_{1}$

Point $P$ Constraints

Substitution for $β$

Solving for $β$

Finding $α$

The locus of the mid points of the chords of the hyperbola $x^{2} - y^{2} = 4$ , which touch the parabola $y^{2} = 8 x$ , is :

If a chord, which is not a tangent, of the parabola $y^{2} = 16 x$ has the equation $2 x + y = p$ , and midpoint $(h, k)$ , then which of the following is(are) possible value(s) of $p, h$ and $k$ ?

Let $P Q$ be a chord of the parabola $y^{2} = 12 x$ and the midpoint of $P Q$ be at $(4, 1)$ . Then, which of the following point lies on the line passing through the points $P$ and $Q$ ?

If $α x + β y = 109$ is the equation of the chord of the ellipse $\frac{x ^{2}}{9} + \frac{y ^{2}}{4} = 1.$ whose mid point is $(\frac{5}{2}, \frac{1}{2})$ then $α + β$ is equal to:

If the midpoint of a chord of the ellipseis $\frac{x ^{2}}{9} + \frac{y ^{2}}{4} = 1$ is $(2, 4/3)$ and the length of the chord is $\frac{2 α}{3}$ then a is:

The length of the chord of the ellipse $\frac{x ^{2}}{4} + \frac{y ^{2}}{2} = 1$ whose mid-point is $(1, \frac{1}{2})$ is:

The equation of the chord, of the ellipse $\frac{x ^{2}}{25} + \frac{y ^{2}}{16} = 1$ whose mid-point is $(3, 1)$ is:

If two distinct chords, drawn from the point $(p, q)$ on the circle $x^{2} + y^{2} = p x + q y$ (where $pq \neq = 0$ ) are bisected by the x-axis, then