Tangents drawn from the point (-8, 0) to the parabola y² = 8x touch the parabola at P and Q. If F is

Q: Wait, why did the $y$ term vanish in the chord of contact equation?

Because the external point $T$ is $(-8, 0)$, so $y_1 = 0$. The formula $yy_1 = 2a(x+x_1)$ becomes $y(0) = 4(x-8)$, which simplifies to $0 = 4(x-8)$, or $x=8$. It is a beautiful simplification that happens because the point lies on the axis of symmetry.

Q: Is there a shortcut for the area of this triangle?

While there are properties involving the coordinates of the points of contact, deriving it from first principles—base and height—is the most robust way to ensure you do not make a sign error. In this case, the base is the vertical segment $PQ$ and the height is the horizontal distance from the focus to the chord.

Q: Why is the height of the triangle $8-2$?

The base $PQ$ is the vertical line $x=8$. The focus $F$ is at $(2, 0)$. The perpendicular distance from the point $(2, 0)$ to the vertical line $x=8$ is simply the difference in their $x$-coordinates: $|8-2| = 6$.

Question

Tangents drawn from the point (-8, 0) to the parabola $y^2 = 8x$ touch the parabola at P and Q. If F is the focus of the parabola, then the area of the triangle PFQ (in sq. units) is equal to :-

Options:

(A) 24

(B) 64

(C) 32

(D) 48

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the point $T(-8, 0)$, looking out at the parabola $y^2 = 8x$. This is not just a math problem; it is a study of how light and geometry interact.

The parabola, defined by $y^2 = 8x$, is a classic curve. By comparing it to the standard form $y^2 = 4ax$, we immediately identify $4a = 8$, which means $a = 2$.

This tells us that the focus $F$ of our parabola is located at $(2, 0)$. This focus is the heart of the parabola, the point where all parallel incoming rays would converge.

### The Chord of Contact: The Bridge

Now, you draw two tangents from $T(-8, 0)$ to the parabola. These tangents touch the curve at two points, $P$ and $Q$. The line segment connecting these points is the 'Chord of Contact.'

In the world of JEE geometry, the formula for the chord of contact from an external point $(x_1, y_1)$ to a parabola $y^2 = 4ax$ is:

$$yy_1 = 2a(x + x_1)$$

By substituting our point $T(-8, 0)$ and our parameter $2a = 4$, we get $y(0) = 4(x - 8)$. The $y$ term vanishes, leaving us with $0 = 4(x - 8)$, which simplifies to $x = 8$.

This is a profound result! It tells us that the chord of contact is a vertical line.

### Finding the Intersection

With the equation of the chord $x = 8$ in hand, finding the coordinates of $P$ and $Q$ becomes a simple task of substitution. We plug $x = 8$ into the parabola equation $y^2 = 8x$:

$$y^2 = 8(8) = 64$$

Thus, $y = \pm 8$. We have found our points: $P(8, 8)$ and $Q(8, -8)$. The symmetry here is beautiful; the points are perfectly mirrored across the $x$-axis.

### The Final Area

We are now looking at a triangle $	riangle PFQ$ with vertices $F(2, 0)$, $P(8, 8)$, and $Q(8, -8)$. To find the area, we treat $PQ$ as the base.

The length of this base is the distance between $y = 8$ and $y = -8$, which is $8 - (-8) = 16$ units. The height of the triangle is the perpendicular distance from the focus $F(2, 0)$ to the vertical line $x = 8$.

This distance is simply $8 - 2 = 6$ units. Using the classic formula for the area of a triangle:

$$	ext{Area} = \frac{1}{2} 	imes 	ext{Base} 	imes 	ext{Height}$$

$$	ext{Area} = \frac{1}{2} 	imes 16 	imes 6 = 48$$

We have arrived at our answer. The area of the triangle is **48 square units**. It is a testament to the elegance of geometry that such a complex-sounding problem resolves into such a clean, integer result.

Animated Solution for Mathematics - Conic Sections: Tangents drawn from the point (-8, 0) to the parabola $y^{2} = 8 x$ touch the parabola at P and Q. If F is the focus of the parabola, then the area of the triangle PFQ (in sq. units) is equal to :-

Select Answer:

Visualized Solution

Analyzing the Parabola $y^{2} = 8 x$

Locating the Focus $F$

External Point and Tangents

Equation of Chord of Contact $P Q$

Substituting Values for $P Q$

Solving for the Chord Equation

Finding Points $P$ and $Q$

Visualizing Triangle $P F Q$

Calculating Base and Height

Final Area Calculation

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

Analyzing the Setup

The Chord of Contact

Finding the Intersection

The Final Area

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