If x + y = k is normal to y² = 12x, then k is (a) 3 (b) 9 (c) -9 (d) -3

Q: Wait, why can't we just use the derivative to find the slope of the tangent and then find the normal?

You absolutely can! If you differentiate $y^2 = 12x$, you get $2y \frac{dy}{dx} = 12$, so the slope of the tangent is $\frac{6}{y}$. The slope of the normal is the negative reciprocal, $-\frac{y}{6}$. By setting this equal to the slope of your line (which is $-1$), you can find the point of contact $(x, y)$ and then find $k$. However, using the standard formula $c = -2am - am^3$ is a much faster, more elegant 'JEE-style' shortcut that saves precious time in the exam hall.

Q: Why is the condition $c = -2am - am^3$ specifically for $y^2 = 4ax$?

Great question! This formula is derived by considering the normal at any point $(at^2, 2at)$ on the parabola. The slope of the tangent at this point is $\frac{1}{t}$, so the slope of the normal is $-t$. If we write the equation of the line as $y = mx + c$, we equate $m = -t$. Substituting $t = -m$ into the normal equation $y = mx - 2am - am^3$ gives us the intercept $c = -2am - am^3$. It's all about connecting the parametric form to the slope-intercept form!

Q: What if the parabola was $x^2 = 4ay$ instead? Would the formula be the same?

No, it would not! The geometry changes because the parabola opens upwards instead of to the right. For $x^2 = 4ay$, the condition for the line $y = mx + c$ to be a normal is different. Always be careful to check the orientation of your parabola before applying the formula. For JEE, it is best to derive it quickly if you forget, or memorize the standard forms for both orientations.

Question

If $x + y = k$ is normal to $y^2 = 12x$, then $k$ is

Options:

(A) 3

(B) 9

(C) -9

(D) -3

Sigma Percentile · Accepted Answer

### The Geometry of Normality

Welcome, fellow traveler on the path to JEE mastery! Today, we are going to peel back the layers of a classic coordinate geometry problem. We are looking at a parabola, $y^2 = 12x$, and a line, $x + y = k$.

Our mission is to find the value of $k$ that forces this line to be a normal to the parabola. This isn't just about plugging numbers into a formula; it is about understanding the rigid, beautiful relationship between a curve and its perpendiculars.

### Phase 1: Decoding the Parabola

Every journey begins with understanding our terrain. We are given the parabola $y^2 = 12x$.

In the language of conic sections, the standard form is $y^2 = 4ax$. By comparing our equation to the standard form, we see that $4a = 12$.

A quick division tells us that $a = 3$. This parameter, $a$, is the heartbeat of our parabola; it tells us exactly how 'wide' the curve is and where its focus lies. With $a = 3$, we have unlocked the first piece of our puzzle.

### Phase 2: The Line's True Identity

Next, we turn our attention to the line $x + y = k$. To understand its role in this geometric dance, we need to see it in its most revealing form: the slope-intercept form, $y = mx + c$.

By rearranging our equation, we get $y = -x + k$. Now, it is crystal clear!

The slope $m$ is $-1$, and the y-intercept $c$ is our mysterious constant $k$. We have successfully translated the line into a form that interacts perfectly with our parabola's properties.

### Phase 3: The Bridge of Normality

Here is where the magic happens. In the world of JEE, we don't always need to derive everything from scratch. We have a powerful tool in our arsenal: the condition for a line $y = mx + c$ to be a normal to the parabola $y^2 = 4ax$.

That condition is given by the following equation:

$$c = -2am - am^3$$

Think of this as the 'bridge' connecting our line's parameters to the parabola's geometry. It is a beautiful, compact expression that encapsulates the requirement of perpendicularity at the point of intersection.

### Phase 4: The Final Calculation

Now, we simply assemble our findings. We have $a = 3$, $m = -1$, and $c = k$. Let's feed these into our bridge equation:

$$k = -2(3)(-1) - 3(-1)^3$$

Let's handle this with care. The first term, $-2(3)(-1)$, simplifies to $6$. The second term, $-3(-1)^3$, requires a moment of focus.

Since $(-1)^3 = -1$, multiplying that by $-3$ gives us $+3$. So, we are left with the final sum:

$$k = 6 + 3 = 9$$

And there it is! The value of **$k$ is $9$**.

The line $x + y = 9$ is not just any line; it is the specific line that strikes the parabola $y^2 = 12x$ at a perfect right angle to the tangent. You have successfully navigated the geometry, identified the parameters, and executed the algebra with precision.

Animated Solution for Mathematics - Conic Sections: If $x + y = k$ is normal to $y^{2} = 12 x$ , then $k$ is

Select Answer:

Visualized Solution

Visualizing the Setup

Standard Form of Parabola

Calculating Parameter $a$

Analyzing the Line Equation

Converting to $y = m x + c$

Identifying $m$ and $c$

Condition for Normality

Substituting the Values

Evaluating the First Term

Evaluating the Second Term

Calculating $k$

Final Conclusion

The Sigma Insight: Equation of Tangent and Normal

The Geometry of Normality

Phase 1

Phase 2

Phase 3

Phase 4

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Animated Solution for Mathematics - Conic Sections: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If x+y=k is normal to y2=12x, then k is