Let ℓ be a line which is normal to the curve y = 2x² + x + 2 at a point P on the curve. If the point

Q: Wait, why can't we just find the tangent at $Q$ and then take the negative reciprocal?

That is a common trap! The point $Q(6, 4)$ lies on the normal line, but it does not necessarily lie on the curve $y = 2x^2 + x + 2$. If you try to find the slope at $Q$ using the curve's derivative, you are finding the slope of the tangent at $Q$ (if $Q$ were on the curve), which is not the slope of the normal line passing through $P$. We must define the normal line based on the unknown point $P$ first.

Q: Why do we use $h$ and $k$ instead of just $x$ and $y$?

Excellent question. We use $h$ and $k$ to distinguish the specific, fixed point $P$ from the general variable $x$ and $y$ that define the curve. By setting $P(h, k)$, we treat $h$ as a constant that we need to solve for. Once we find $h$, the entire geometry of the normal line is fixed.

Q: Is there only one point $P$ that satisfies this condition?

In this specific problem, the cubic equation $4h^3 + 3h^2 - 3h - 4 = 0$ might have multiple roots. However, for the purpose of finding the area of the triangle, we look for a valid coordinate $P$. Often, in JEE problems, one root is easily found by inspection (like $h=1$ here), and that root provides the intended point $P$.

Question

Let $\ell$ be a line which is normal to the curve $y = 2x^2 + x + 2$ at a point P on the curve. If the point Q(6, 4) lies on the line $\ell$ and O is origin, then the area of the triangle OPQ is equal to ________.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a roller coaster track defined by the parabolic path $y = 2x^2 + x + 2$. You are at a specific point $P(h, k)$, and you want to drop a plumb line—a normal—straight down from your position.

This normal line is perpendicular to the track at your exact location. We are told that this line, which we call $\ell$, passes through a distant point $Q(6, 4)$. Our mission is to find the area of the triangle formed by the origin $O(0, 0)$, your position $P$, and the point $Q$.

### The Tangent and the Normal

To understand the normal, we must first understand the tangent. The slope of the tangent at any point $x$ on our curve is the derivative:

$$\frac{dy}{dx} = 4x + 1$$

At our specific point $P(h, k)$, the slope of the tangent is $m_t = 4h + 1$. The normal is the line perpendicular to the tangent, so its slope $m_n$ is the negative reciprocal:

$$m_n = -\frac{1}{4h + 1}$$

### The Algebraic Bridge

Using the point-slope form, the equation of our normal line $\ell$ is $y - k = m_n(x - h)$. We know that the point $Q(6, 4)$ lies on this line, so substituting these coordinates gives:

$$4 - k = -\frac{1}{4h + 1}(6 - h)$$

Since point $P(h, k)$ lies on the curve $y = 2x^2 + x + 2$, we have the constraint $k = 2h^2 + h + 2$. Substituting this expression for $k$ into our normal equation allows us to solve for $h$.

### The Cubic Challenge

Substituting $k$ into the normal equation yields $4 - (2h^2 + h + 2) = -\frac{1}{4h + 1}(6 - h)$. Simplifying the left side results in $2 - h - 2h^2$.

Multiplying both sides by $(4h + 1)$ and rearranging terms, we arrive at the following cubic equation:

$$4h^3 + 3h^2 - 3h - 4 = 0$$

By testing simple integer values, we find that $h = 1$ satisfies the equation: $4(1)^3 + 3(1)^2 - 3(1) - 4 = 4 + 3 - 3 - 4 = 0$. With $h = 1$, we calculate $k = 2(1)^2 + 1 + 2 = 5$. Thus, our point $P$ is $(1, 5)$.

### Final Calculation

We now have the vertices of our triangle: $O(0, 0)$, $P(1, 5)$, and $Q(6, 4)$. The area of a triangle with one vertex at the origin is given by the formula:

$$	ext{Area} = \frac{1}{2} |x_1y_2 - x_2y_1|$$

Substituting our coordinates into the formula:

$$	ext{Area} = \frac{1}{2} |(1)(4) - (6)(5)| = \frac{1}{2} |4 - 30| = \frac{1}{2} |-26| = 13$$

The final area of the triangle is **13 square units**.

Animated Solution for Mathematics - Differentiation: Let $ℓ$ be a line which is normal to the curve $y = 2 x^{2} + x + 2$ at a point P on the curve. If the point Q(6, 4) lies on the line $ℓ$ and O is origin, then the area of the triangle OPQ is equal to ________.

Enter Numerical Value:

Visualized Solution

Visualize the Curve and Point P

Find the Slope of the Tangent

Determine the Slope of the Normal

Formulate the Normal Equation

Substitute Point Q

Relate h and k using the Curve

Solve the Algebraic Equation

Find the Roots

Find Coordinates of Point P

Visualize Triangle OPQ

Area Formula for Triangle

Calculate the Area

Final Answer

The Sigma Insight: Tangents, Normals and Rate Measure

Analyzing the Setup

The Tangent and the Normal

The Algebraic Bridge

The Cubic Challenge

Final Calculation

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