Let the length of the focal chord PQ of the parabola y² = 12x be 15 units. If the distance of PQ fro

Q: Why do we use the $\csc^2 \theta$ formula instead of parametric coordinates $(at_1^2, 2at_1)$?

Parametric coordinates are powerful, but for focal chords, the angular approach is often faster. It bypasses the need to solve for two separate parameters $t_1$ and $t_2$, allowing you to jump straight to the slope and the line equation.

Q: Does the sign of the slope $m = \pm 2$ affect the final distance $p$?

Great observation! Since the distance formula involves an absolute value and the slope is squared in the denominator, the sign of the slope does not change the distance. The geometry is symmetric, so both slopes yield the same distance $p$.

Q: Is there a shortcut for the distance of a focal chord from the origin?

Yes, for a parabola $y^2 = 4ax$, the distance $p$ from the origin to a focal chord with angle $\theta$ is $p = a \sin \theta \cos \theta$. However, deriving it from the line equation is safer and less prone to memory errors during the exam.

Question

Let the length of the focal chord $PQ$ of the parabola $y^2 = 12x$ be 15 units. If the distance of $PQ$ from the origin is $p$, then $10p^2$ is equal to _______

Sigma Percentile · Accepted Answer

### Analyzing the Setup

When you look at the equation $y^2 = 12x$, I want you to see more than just algebra. I want you to see a curve that is defined by its focus and its directrix.

By comparing this to the standard form $y^2 = 4ax$, we immediately identify $4a = 12$, which gives us $a = 3$. This tells us that the focus $S$ sits at $(3, 0)$.

This point is the heartbeat of our parabola. Every focal chord must pass through this point. It is the anchor around which our geometry revolves.

### The Power of the Angular Approach

Now, we are given a focal chord $PQ$ with a length of $15$ units. You could use parametric coordinates, but there is a more sophisticated path.

The length of a focal chord making an angle $	heta$ with the axis of the parabola is given by the beautiful relation:

$$L = 4a \csc^2 	heta$$

Substituting our known values, we get $15 = 12 \csc^2 	heta$. This simplifies to:

$$\csc^2 	heta = \frac{15}{12} = \frac{5}{4}$$

By taking the reciprocal, we find that $\sin^2 	heta = \frac{4}{5}$. We are now standing on the threshold of finding the slope of this line.

### Bridging Trigonometry and Algebra

We need the slope $m = 	an 	heta$ to define the line. We know that $	an^2 	heta = \frac{\sin^2 	heta}{1 - \sin^2 	heta}$.

Substituting our value, we get:

$$	an^2 	heta = \frac{\frac{4}{5}}{1 - \frac{4}{5}} = \frac{\frac{4}{5}}{\frac{1}{5}} = 4$$

This gives us $m = \pm 2$. Whether the slope is $2$ or $-2$ does not matter for the distance calculation, as the geometry is symmetric.

Let us proceed with $m = 2$. The line passes through the focus $(3, 0)$ with a slope of $2$. Using the point-slope form, $y - 0 = 2(x - 3)$, we arrive at the equation $2x - y - 6 = 0$.

### The Final Distance

Finally, we calculate the perpendicular distance $p$ from the origin $(0, 0)$ to this line. The formula is:

$$p = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$$

Plugging in our values, we get:

$$p = \frac{|2(0) - 1(0) - 6|}{\sqrt{2^2 + (-1)^2}} = \frac{|-6|}{\sqrt{5}} = \frac{6}{\sqrt{5}}$$

The problem asks for $10p^2$. Squaring $p$, we get $p^2 = \frac{36}{5}$.

Multiplying by $10$, we have $10 	imes \frac{36}{5} = 2 	imes 36 = 72$.

The final answer is **72**.

Animated Solution for Mathematics - Conic Sections: Let the length of the focal chord $P Q$ of the parabola $y^{2} = 12 x$ be 15 units. If the distance of $P Q$ from the origin is $p$ , then $10 p^{2}$ is equal to _______

Enter Numerical Value:

Visualized Solution

Analyze the Parabola $y^{2} = 12 x$

Recall Focal Chord Length Formula

Solve for $sin^{2} θ$

Determine the Slope $tan θ$

Equation of the Focal Chord

Distance $p$ from Origin

Calculate the Value of $p$

Final Computation of $10 p^{2}$

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

The Power of the Angular Approach

Bridging Trigonometry and Algebra

The Final Distance

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Enter Numerical Value:

Visualized Solution

Analyze the Parabola .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y2=12x

Recall Focal Chord Length Formula

Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sin2θ

Determine the Slope .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tanθ