JEE Advanced 2002

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Animated Solution for Mathematics - Differentiation: The point(s) on the curve $y^{3} + 3 x^{2} = 12 y$ where the tangent is vertical, is (are)

Q: Wait, why do we set the denominator to zero? Shouldn't we set the whole derivative to zero for a tangent?

That is a common confusion! Setting the derivative $\frac{dy}{dx}$ to zero gives you a horizontal tangent, where the line is flat. For a vertical tangent, the slope is undefined, which mathematically happens when the denominator of the fraction $\frac{dy}{dx}$ is zero, causing the value to blow up to infinity.

Select Answer:

Progressive Hints

Hint 1

Hint 1: Visualize the geometry of the curve. A vertical tangent occurs where the tangent line is parallel to the y-axis, meaning the slope of the tangent, denoted as

\frac{d y}{d x}

, becomes undefined or approaches infinity. This happens when the denominator of the derivative expression is zero.

Hint 2

Hint 2: Use implicit differentiation on the equation

y^{3} + 3 x^{2} = 12 y

. Differentiate both sides with respect to

x

, remembering to apply the chain rule for terms involving

y

. Your goal is to isolate

\frac{d y}{d x}

and express it as a function of

x

and

y

Hint 3

Hint 3: After finding

\frac{d y}{d x} = \frac{- 2 x}{y ^{2} - 4}

, set the denominator

y^{2} - 4 = 0

to find the potential

y

-coordinates. Then, substitute these

y

-values back into the original curve equation to find the corresponding

x

-coordinates, ensuring you reject any cases that do not yield real solutions for

x

Visualized Solution

Visualizing the Curve

Given curve: $y^{3} + 3 x^{2} = 12 y$
Objective: Find points where the tangent is vertical.
A vertical tangent implies the slope $\frac{d y}{d x}$ is undefined (infinite).

Implicit Differentiation

Differentiating both sides with respect to $x$ :
$\frac{d}{d x} (y^{3}) + \frac{d}{d x} (3 x^{2}) = \frac{d}{d x} (12 y)$

Applying the Chain Rule

$3 y^{2} \frac{d y}{d x} + 6 x = 12 \frac{d y}{d x}$

Grouping $\frac{d y}{d x}$ Terms

Group $\frac{d y}{d x}$ terms on one side:
$3 y^{2} \frac{d y}{d x} - 12 \frac{d y}{d x} = - 6 x$
$(3 y^{2} - 12) \frac{d y}{d x} = - 6 x$

Isolating the Slope $\frac{d y}{d x}$

$\frac{d y}{d x} = \frac{- 6 x}{3 y ^{2} - 12}$
Simplify by dividing numerator and denominator by $3$ :
$\frac{d y}{d x} = \frac{- 2 x}{y ^{2} - 4}$

Condition for Vertical Tangent

For a vertical tangent, the denominator of $\frac{d y}{d x}$ must be zero.
$y^{2} - 4 = 0$

Solving for $y$

$y^{2} = 4$
Possible $y$ coordinates: $y = 2$ or $y = - 2$ .

Case 1: $y = 2$

Substitute $y = 2$ into the original curve equation:
$(2)^{3} + 3 x^{2} = 12 (2)$

Solving for $x$

$8 + 3 x^{2} = 24$
$3 x^{2} = 16$

First Set of Points

$x^{2} = \frac{16}{3}$
$x = \pm \frac{4}{3}$
Points: $(\pm \frac{4}{3}, 2)$

Case 2: $y = - 2$

Substitute $y = - 2$ into the original curve equation:
$(- 2)^{3} + 3 x^{2} = 12 (- 2)$

Analyzing the Result

$- 8 + 3 x^{2} = - 24$
$3 x^{2} = - 16$

Rejecting the Invalid Case

$x^{2} = - \frac{16}{3}$
Since $x^{2}$ cannot be negative for real $x$ , there are no real solutions for $y = - 2$ .
This case is rejected.

Final Conclusion

The only points on the curve with a vertical tangent are $(\pm \frac{4}{3}, 2)$ .
Correct Option: (4)

00:00 / 00:00

The Sigma Insight: Tangents, Normals and Rate Measure

Analyzing the Setup

The path is defined by the implicit equation:

y^{3} + 3 x^{2} = 12 y

We seek the points where the tangent line is perfectly vertical. Geometrically, this occurs where the slope

\frac{d y}{d x}

becomes undefined, which happens when the denominator of the derivative expression is zero while the numerator remains non-zero.

The Dance of Implicit Differentiation

To find the slope, we differentiate both sides of the equation with respect to

x

\frac{d}{d x} (y^{3}) + \frac{d}{d x} (3 x^{2}) = \frac{d}{d x} (12 y)

Applying the chain rule, we obtain:

3 y^{2} \frac{d y}{d x} + 6 x = 12 \frac{d y}{d x}

Isolating the Slope

We group the terms containing

\frac{d y}{d x}

on one side to isolate the derivative:

3 y^{2} \frac{d y}{d x} - 12 \frac{d y}{d x} = - 6 x

Factoring out

\frac{d y}{d x}

, we get:

(3 y^{2} - 12) \frac{d y}{d x} = - 6 x

Dividing by the coefficient of

\frac{d y}{d x}

and simplifying by a factor of

3

, we arrive at the general expression for the slope:

\frac{d y}{d x} = \frac{- 2 x}{y ^{2} - 4}

The Moment of Infinity

A vertical tangent occurs when the denominator equals zero:

y^{2} - 4 = 0 \Rightarrow y = \pm 2

We must verify which of these

y

-values correspond to real points on the original curve.

Case 1: $y = 2$ Substituting into the original equation:

(2)^{3} + 3 x^{2} = 12 (2)

8 + 3 x^{2} = 24 \Rightarrow 3 x^{2} = 16 \Rightarrow x = \pm \frac{4}{3}

These yield the valid points:

(\pm \frac{4}{3}, 2)

Case 2: $y = - 2$ Substituting into the original equation:

(- 2)^{3} + 3 x^{2} = 12 (- 2)

- 8 + 3 x^{2} = - 24 \Rightarrow 3 x^{2} = - 16

Since

x^{2}

cannot be negative for real

x

, this case provides no real solutions.

Final Conclusion

The points on the curve where the tangent line is vertical are:

(\frac{4}{3}, 2) and (- \frac{4}{3}, 2)

Common Traps & FAQs

Wait, why do we set the denominator to zero? Shouldn't we set the whole derivative to zero for a tangent?

That is a common confusion! Setting the derivative

\frac{d y}{d x}

to zero gives you a horizontal tangent, where the line is flat. For a vertical tangent, the slope is undefined, which mathematically happens when the denominator of the fraction

\frac{d y}{d x}

is zero, causing the value to blow up to infinity.

Why did we reject the case where

y = - 2

? Is it just because the math didn't work out?

It is not just about the math not working; it is about the physical reality of the curve. When we solved for

x

in the case

y = - 2

, we got

3 x^{2} = - 16

, which implies

x^{2} = - \frac{16}{3}

. Since the square of any real number must be non-negative, there is no real

x

that satisfies this. Therefore, that point simply does not exist on the curve in the real coordinate plane.

Could I have solved this by finding

\frac{d x}{d y}

instead?

Absolutely! In fact, that is a brilliant shortcut. If you find

\frac{d x}{d y}

and set it to zero, you are directly finding the points where the tangent is vertical. It often saves time and avoids the potential for algebraic errors when isolating

\frac{d y}{d x}

. Both methods are valid, but choosing the one that aligns with the condition (vertical vs. horizontal) is a sign of a seasoned problem solver.

Animated Solution for Mathematics - Differentiation: The point(s) on the curve $y^{3} + 3 x^{2} = 12 y$ where the tangent is vertical, is (are)

Select Answer:

Visualized Solution

Visualizing the Curve

Implicit Differentiation

Applying the Chain Rule

Grouping $\frac{d y}{d x}$ Terms

Isolating the Slope $\frac{d y}{d x}$

Condition for Vertical Tangent

Solving for $y$

Case 1: $y = 2$

Solving for $x$

First Set of Points

Case 2: $y = - 2$

Analyzing the Result

Rejecting the Invalid Case

Final Conclusion

The Sigma Insight: Tangents, Normals and Rate Measure

Analyzing the Setup

The Dance of Implicit Differentiation

Isolating the Slope

The Moment of Infinity

Final Conclusion

Similar Questions

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If the angle made by the tangent at the point $(x_{0}, y_{0})$ on the curve $x = 12 (t + sin t cos t)$ , $y = 12 (1 + sin t)^{2}, 0 < t < \frac{π}{2}$ , with the positive x-axis is $\frac{π}{3}$ , then $y_{0}$ is equal to

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The equation of the tangent to the curve $y = x + \frac{4}{x ^{2}}$ , that is parallel to the $x$ -axis, is

the tangent at the point $(x_{1}, y_{1})$ on the curve $y = x^{3} + 3 x^{2} + 5$ passes through the origin, then $(x_{1}, y_{1})$ does NOT lie on the curve :

Find the equation of the normal to the curve $y = (1 + x)^{y} + sin^{- 1} (sin^{2} x)$ at $x = 0$

The equation of the normal to the curve $y = (1 + x)^{2 y} + cos^{2} (sin^{- 1} x)$ at $x = 0$ is:

Let the normal at a point $P$ on the curve $y^{2} - 3 x^{2} + y + 10 = 0$ intersect the $y$ -axis at $(0, \frac{3}{2})$ . If $m$ is the slope of the tangent at $P$ to the curve, then $∣ m ∣$ is equal to

If the tangent to the curve $y = \frac{x}{x ^{2} - 3}, x \in R, x \neq = \pm 3$ , at a point $(α, β) \neq = (0, 0)$ on it is parallel to the line $2 x + 6 y - 11 = 0$ , then :

Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The point(s) on the curve y3+3x2=12y where the tangent is vertical, is (are)

Select Answer:

Visualized Solution

Visualizing the Curve

Implicit Differentiation

Applying the Chain Rule

Grouping .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dxdy​ Terms

Isolating the Slope .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dxdy​

Condition for Vertical Tangent

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y

Case 1: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=2

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

First Set of Points

Case 2: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=−2

Analyzing the Result

Rejecting the Invalid Case

Final Conclusion

The Sigma Insight: Tangents, Normals and Rate Measure

Analyzing the Setup

The Dance of Implicit Differentiation

Isolating the Slope

The Moment of Infinity

Final Conclusion

Similar Questions

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Animated Solution for Mathematics - Differentiation: The point(s) on the curve $y^{3} + 3 x^{2} = 12 y$ where the tangent is vertical, is (are)

Grouping $\frac{d y}{d x}$ Terms

Isolating the Slope $\frac{d y}{d x}$

Solving for $y$

Case 1: $y = 2$

Solving for $x$

Case 2: $y = - 2$

Which of the following points lies on the tangent to the curve $x^{4} e^{y} + 2 y + 1 = 3$ at the point $(1, 0)$ ?

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:

The equation of the tangent to the curve $y = x + \frac{4}{x ^{2}}$ , that is parallel to the $x$ -axis, is

the tangent at the point $(x_{1}, y_{1})$ on the curve $y = x^{3} + 3 x^{2} + 5$ passes through the origin, then $(x_{1}, y_{1})$ does NOT lie on the curve :

Find the equation of the normal to the curve $y = (1 + x)^{y} + sin^{- 1} (sin^{2} x)$ at $x = 0$

The equation of the normal to the curve $y = (1 + x)^{2 y} + cos^{2} (sin^{- 1} x)$ at $x = 0$ is:

Let the normal at a point $P$ on the curve $y^{2} - 3 x^{2} + y + 10 = 0$ intersect the $y$ -axis at $(0, \frac{3}{2})$ . If $m$ is the slope of the tangent at $P$ to the curve, then $∣ m ∣$ is equal to

If the tangent to the curve $y = \frac{x}{x ^{2} - 3}, x \in R, x \neq = \pm 3$ , at a point $(α, β) \neq = (0, 0)$ on it is parallel to the line $2 x + 6 y - 11 = 0$ , then :