A tangent and a normal are drawn at the point P(2, -4) on the parabola y² = 8x, which meet the direc

Q: Wait, why is the directrix $x = -2$? I thought it was $x = -a$.

You are absolutely right! The standard form of a parabola opening to the right is $y^2 = 4ax$. By comparing $y^2 = 8x$ with $y^2 = 4ax$, we find $4a = 8$, which means $a = 2$. The directrix for this standard form is $x = -a$. Substituting $a = 2$, we get $x = -2$. It is a simple step, but always double-check your value of $a$!

Q: Why do we assume the diagonals of the square $AQBP$ bisect each other?

That is a great question! This is a universal property of all parallelograms, and since a square is a special type of parallelogram, its diagonals must bisect each other. This means the midpoint of $AB$ is the same as the midpoint of $PQ$. It is a powerful shortcut that saves us from having to calculate the lengths of the sides!

Q: Is there a faster way to find the equations of the tangent and normal?

For the tangent, the $T = 0$ method is the most reliable. For the normal, once you have the slope of the tangent ($m_T$), the slope of the normal is simply $-1/m_T$ because they are perpendicular. Using the point-slope form $y - y_1 = m_N(x - x_1)$ is the most efficient path. Keep practicing these, and they will become second nature!

Question

A tangent and a normal are drawn at the point $P(2, -4)$ on the parabola $y^2 = 8x$, which meet the directrix of the parabola at the points $A$ and $B$ respectively. If $Q(a, b)$ is a point such that $AQBP$ is a square, then $2a + b$ is equal to :

Options:

(A) $-16$

(B) $-18$

(C) $-12$

(D) $-20$

Sigma Percentile · Accepted Answer

### The Geometry of a Square: A Parabolic Journey

Welcome, future engineer! Today, we are not just solving a problem; we are exploring the elegant dance between a parabola, its tangent, its normal, and a square.

Coordinate geometry is often seen as a collection of formulas, but I want you to see it as a landscape. Imagine you are standing on the Cartesian plane, looking at the parabola defined by $y^2 = 8x$. This curve is our stage.

### Phase 1: The Anatomy of the Parabola

Every parabola has a heartbeat—its focus and its directrix. To find them, we compare our given equation, $y^2 = 8x$, with the standard form $y^2 = 4ax$.

By inspection, we see that $4a = 8$, which gives us $a = 2$. The directrix, the line that acts as the 'mirror' for the parabola, is defined by the equation $x = -a$.

Thus, our directrix is the vertical line $x = -2$. Keep this line in your mind; it is the destination for our tangent and normal.

### Phase 2: The Tangent and Normal at P

We are given a point $P(2, -4)$ on this parabola. To find the tangent at this point, we use the elegant $T = 0$ method.

For a parabola $y^2 = 4ax$, the tangent at $(x_1, y_1)$ is $yy_1 = 2a(x + x_1)$. Substituting our values, we get $y(-4) = 4(x + 2)$.

Simplifying this, we arrive at the beautiful, simple equation:
$$x + y + 2 = 0$$

Now, what about the normal? The normal is the line perpendicular to the tangent at the point of contact.

The slope of our tangent, $x + y + 2 = 0$, is $m_T = -1$. Since the normal is perpendicular, its slope $m_N$ must be the negative reciprocal, which is $m_N = 1$.

Using the point-slope form $y - y_1 = m_N(x - x_1)$, we get $y - (-4) = 1(x - 2)$, which simplifies to:
$$x - y - 6 = 0$$

### Phase 3: The Intersection

Our tangent and normal are now racing toward the directrix $x = -2$. Let's find where they land.

For the tangent, we substitute $x = -2$ into $x + y + 2 = 0$, giving us $-2 + y + 2 = 0$, so $y = 0$. Point $A$ is $(-2, 0)$.

For the normal, we substitute $x = -2$ into $x - y - 6 = 0$, giving us $-2 - y - 6 = 0$, so $y = -8$. Point $B$ is $(-2, -8)$.

### Phase 4: The Square AQBP

Here is where the magic happens. We are told that $AQBP$ forms a square. In any square, the diagonals bisect each other.

This means the midpoint of diagonal $AB$ must be the same as the midpoint of diagonal $PQ$. Let's find the midpoint of $AB$:
$$\left( \frac{-2 + (-2)}{2}, \frac{0 + (-8)}{2} \right) = (-2, -4)$$

Now, let $Q$ be $(a, b)$. The midpoint of $PQ$ is $\left( \frac{2 + a}{2}, \frac{-4 + b}{2} \right)$. Equating this to $(-2, -4)$, we get:

1. $\frac{2 + a}{2} = -2 \Rightarrow 2 + a = -4 \Rightarrow a = -6$

2. $\frac{-4 + b}{2} = -4 \Rightarrow -4 + b = -8 \Rightarrow b = -4$

### The Final Step

We have found $a = -6$ and $b = -4$. The question asks for the value of $2a + b$.

Substituting our values, we get:
$$2(-6) + (-4) = -12 - 4 = -16$$

And there it is! Through careful visualization and the application of geometric properties, we have arrived at the solution. The final answer is **-16**.

Select Answer:

Visualized Solution

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

Tangent at Point $P (2, - 4)$

Find Intersection Point $A$

Normal at Point $P (2, - 4)$

Find Intersection Point $B$

Properties of Square $A QB P$

Calculate Midpoint of $A B$

Solve for $Q (a, b)$

Final Calculation: $2 a + b$

The Sigma Insight: Equation of Tangent and Normal

The Geometry of a Square

Phase 1

Phase 2

Phase 3

Phase 4

The Final Step

Similar Questions

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Normals are drawn from the point $P$ with slopes $m_{1}, m_{2}, m_{3}$ to the parabola $y^{2} = 4 x$ . If locus of $P$ with $m_{1} m_{2} = α$ is a part of the parabola itself then find $α$ .

Select Answer:

Visualized Solution

Analyze the Parabola .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y2=8x

Tangent at Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(2,−4)

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

Normal at Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(2,−4)

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

Properties of Square .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AQBP

Calculate Midpoint of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AB

Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q(a,b)

Final Calculation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2a+b