Let P(h, k) be a point on the curve y = x² + 7x + 2, nearest to the line, y = 3x - 3 Then the equati

Q: Why do we assume the tangent is parallel to the line? Can't we just use the distance formula?

You absolutely could use the distance formula, but that would involve minimizing a function of two variables, which is algebraically tedious. The geometric insight—that the closest point on a curve to a line occurs where the tangent is parallel to that line—is a powerful shortcut that saves you time and reduces the chance of calculation errors. It is the 'JEE way' of thinking!

Q: Why is the slope of the normal $-\frac{1}{3}$?

Great question! The normal is defined as the line perpendicular to the tangent at the point of contact. In coordinate geometry, if two lines are perpendicular, the product of their slopes is $-1$. Since our tangent slope is $3$, we have $3 \times m_n = -1$, which gives $m_n = -\frac{1}{3}$.

Q: I got $x = -2$, but how do I make sure I did not make a sign error in the final equation?

It is easy to trip on signs! Always double-check your point-slope substitution. You have $y - (-8) = -\frac{1}{3}(x - (-2))$. This becomes $y + 8 = -\frac{1}{3}(x + 2)$. Multiplying by $3$ gives $3y + 24 = -x - 2$. Moving everything to one side gives $x + 3y + 26 = 0$. Take it slow, one step at a time!

Question

Let $P(h, k)$ be a point on the curve $y = x^2 + 7x + 2$, nearest to the line, $y = 3x - 3$. Then the equation of the normal to the curve at $P$ is:

Options:

(A) $x - 3y - 11 = 0$

(B) $x - 3y + 22 = 0$

(C) $x + 3y - 62 = 0$

(D) $x + 3y + 26 = 0$

Sigma Percentile · Accepted Answer

### The Geometry of Proximity: A Journey to the Nearest Point

Welcome, future engineers! Today, we are going to explore a beautiful problem that bridges the gap between pure geometry and the power of calculus.

We are tasked with finding a point $P(h, k)$ on the parabola $y = x^2 + 7x + 2$ that sits closest to the line $y = 3x - 3$. It is a classic JEE trap that tests your ability to visualize the relationship between curves and lines.

### Phase 1: The Geometric Intuition

Imagine you are standing on a vast, flat plane. You have a curved path, a parabola, and a straight road, the line $y = 3x - 3$.

If you were to slide the road parallel to itself, moving it closer and closer to the path, the very first point where the road touches the path is the point of minimum distance.

At this exact moment of contact, the road (the line) and the path (the curve) share the same direction. In mathematical terms, the tangent to the curve at that point $P$ must be parallel to the given line. This is our golden key!

### Phase 2: The Calculus Bridge

Since the tangent at $P$ is parallel to the line $y = 3x - 3$, their slopes must be identical. Comparing the line to the standard form $y = mx + c$, we see the slope $m = 3$.

Now, we need to find the slope of the curve at any point. We differentiate the curve equation $y = x^2 + 7x + 2$ with respect to $x$ to get the slope function:

$$ \frac{dy}{dx} = 2x + 7 $$

We set this derivative equal to the slope of our line:

$$ 2x + 7 = 3 $$

Solving this, we find $2x = -4$, which gives us $x = -2$. We have successfully located the $x$-coordinate of our point $P$!

### Phase 3: The Point of Contact

Now that we have the $x$-coordinate, finding the $y$-coordinate is straightforward. We substitute $x = -2$ back into the original curve equation:

$$ y = (-2)^2 + 7(-2) + 2 $$

Calculating this, we get $y = 4 - 14 + 2 = -8$. So, our point of contact $P$ is $(-2, -8)$.

### Phase 4: The Normal Line

The question asks for the equation of the normal at $P$. The normal is the line perpendicular to the tangent.

If the tangent has a slope $m_t = 3$, then the normal must have a slope $m_n$ defined by:

$$ m_n = -\frac{1}{m_t} = -\frac{1}{3} $$

Now we have everything we need: a point $P(-2, -8)$ and a slope $m_n = -\frac{1}{3}$. Using the point-slope form $y - y_1 = m_n(x - x_1)$, we write:

$$ y - (-8) = -\frac{1}{3}(x - (-2)) $$

### The Final Elegance

Let us simplify this carefully. The equation becomes $y + 8 = -\frac{1}{3}(x + 2)$.

Multiplying both sides by $3$, we get $3(y + 8) = -(x + 2)$, which simplifies to $3y + 24 = -x - 2$.

Rearranging all terms to one side, we arrive at the final, elegant equation:

**$x + 3y + 26 = 0$**

You have just navigated the geometry of the parabola and the calculus of slopes to find the exact normal. Keep this logic in your toolkit—it is the foundation for solving many complex problems in coordinate geometry!

Animated Solution for Mathematics - Differentiation: Let $P (h, k)$ be a point on the curve $y = x^{2} + 7 x + 2$ , nearest to the line, $y = 3 x - 3$ . Then the equation of the normal to the curve at $P$ is:

Select Answer:

Visualized Solution

Visualizing the Curve and Line

The Nearest Point Condition

Slope of the Tangent

Differentiating the Curve

Finding the $x$ -coordinate of $P$

Finding the $y$ -coordinate of $P$

Slope of the Normal

Equation of the Normal

Final Simplification

The Sigma Insight: Tangents, Normals and Rate Measure

The Geometry of Proximity

Phase 1

Phase 2

Phase 3

Phase 4

The Final Elegance

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