The normal to the curve y(x – 2)(x – 3) = x + 6 at the point where the curve intersects the y–axis p

Q: Why do we use the product rule instead of expanding the brackets first?

You absolutely could expand the brackets! It is a valid path. However, in JEE, time is your most precious resource. Using the product rule directly on $y(x-2)(x-3)$ is often faster and reduces the risk of algebraic errors that can occur during expansion. Both methods will lead you to the same result, so choose the one where you feel most confident.

Q: Wait, why is the normal slope the negative reciprocal of the tangent slope?

That is a fundamental geometric property! If a line has a slope $m_t$, any line perpendicular to it must have a slope $m_n$ such that $m_t \cdot m_n = -1$. This ensures the lines are at a $90^{\circ}$ angle to each other. It is the core definition of a normal line.

Q: Is there a trick to finding the y-intercept?

There is no 'trick' per se, but there is a golden rule: the y-axis is defined by the equation $x=0$. Whenever you see 'intersects the y-axis', immediately substitute $x=0$ into the equation. It is the most reliable way to find the point of intersection.

Question

The normal to the curve $y(x - 2)(x - 3) = x + 6$ at the point where the curve intersects the y-axis passes through the point:

Options:

(A) $(-\frac{1}{2}, -\frac{1}{2})$

(B) $(\frac{1}{2}, \frac{1}{2})$

(C) $(-\frac{1}{2}, \frac{1}{3})$

(D) $(\frac{1}{2}, \frac{1}{3})$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a vast coordinate plane, looking at a curve defined by the equation $y(x-2)(x-3) = x+6$. Every curve has a story to tell, and every point on it has a unique character.

Today, we are going to uncover the secret of this curve at the exact moment it crosses the y-axis. We are not just solving for a line; we are finding the normal—the perpendicular guardian of the tangent at that point.

### The Hunt for the Point

Our journey begins with a simple, elegant truth. Whenever a curve intersects the y-axis, the x-coordinate is always zero. This is our anchor.

By substituting $x=0$ into our original equation:

$$y(0-2)(0-3) = 0+6$$

The complexity melts away. We are left with $y(-2)(-3) = 6$, which simplifies to $6y = 6$, and finally, $y=1$. We have found our point of interest: $P(0, 1)$.

### The Calculus of Change

Now, we need to understand how the curve behaves at $P$. To do this, we need the slope of the tangent. We must differentiate the equation $y(x-2)(x-3) = x+6$ with respect to $x$.

We use the Product Rule: $\frac{d}{dx}[u \cdot v] = u'v + uv'$. Let $u = y$ and $v = (x-2)(x-3) = x^2 - 5x + 6$.

When we differentiate, we get:

$$y'(x^2 - 5x + 6) + y(2x - 5) = 1$$

This equation captures the rate of change of the curve at any point $x$. By substituting our point $P(0, 1)$ into this derivative, we find:

$$y'(0-0+6) + 1(0-5) = 1$$

This simplifies to $6y' - 5 = 1$, or $6y' = 6$, giving us a slope of $m_t = 1$. The tangent is rising at a perfect 45-degree angle.

### The Geometry of the Normal

We have the tangent, but the problem asks for the normal. The normal is the line perpendicular to the tangent at the point of contact.

If the tangent's slope is $m_t = 1$, the normal's slope $m_n$ must be the negative reciprocal:

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

Now, we have everything we need: a point $(0, 1)$ and a slope $m_n = -1$. Using the point-slope form, $y - y_1 = m_n(x - x_1)$, we get:

$$y - 1 = -1(x - 0)$$

This simplifies beautifully to **$x + y = 1$**.

### The Final Verification

In the world of JEE, finding the equation is only half the battle. We must now verify which of the given options lies on this line.

We test the point $(\frac{1}{2}, \frac{1}{2})$. Substituting these into our equation $x + y = 1$, we get:

$$\frac{1}{2} + \frac{1}{2} = 1$$

It is a perfect match! You have successfully navigated the curve, mastered the calculus, and verified the result.

Animated Solution for Mathematics - Differentiation: The normal to the curve $y (x - 2) (x - 3) = x + 6$ at the point where the curve intersects the y-axis passes through the point:

Select Answer:

Visualized Solution

Analyze the Curve Equation

Condition for y-intercept

Calculate the y-intercept

Differentiate the Equation

Apply Product Rule

Substitute Point P

Calculate Tangent Slope ( $m_{t}$ )

Find Normal Slope ( $m_{n}$ )

Equation of the Normal

Verify the Given Options

Final Conclusion

The Sigma Insight: Tangents, Normals and Rate Measure

Analyzing the Setup

The Hunt for the Point

The Calculus of Change

The Geometry of the Normal

The Final Verification

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Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The normal to the curve y(x−2)(x−3)=x+6 at the point where the curve intersects the y-axis passes through the point: