JEE Advanced 2001

LEVELJEE Main

Animated Solution for Mathematics - Differentiation: The triangle formed by the tangent to the curve $f (x) = x^{2} + b x - b$ at the point $(1, 1)$ and the coordinate axes, lies in the first quadrant. If its area is $2$ , then the value of $b$ is

Q: Wait, why do we need to check the first quadrant condition? Can't we just solve the area equation?

That is a great question! While the algebra might give you a value for $b$, the geometry of the problem is the master. The area formula $Area = \frac{1}{2} \times base \times height$ assumes that the base and height are positive lengths. If the intercepts were negative, the triangle would not be in the first quadrant, and the area calculation would be physically meaningless in this context. Always check your constraints!

Q: Why can't we just use energy conservation or some other physics principle?

I love the enthusiasm for physics! However, this is a pure coordinate geometry and calculus problem. We are dealing with the properties of a curve and its tangent line. The 'energy' here is the mathematical consistency of the tangent equation. Stick to the calculus toolkit—derivatives and line equations—and you will find the path.

Q: What if the quadratic equation had two distinct roots? How would I know which one to pick?

If you had two roots, you would test each one against the constraints we derived from the first-quadrant condition (i.e., $b < -2$). If one root satisfies the condition and the other does not, you reject the invalid one. In this specific problem, the quadratic is a perfect square, which simplifies things, but always be ready to filter your solutions!

Select Answer:

Progressive Hints

Hint 1

The Spark: Visualize the tangent line as a geometric entity. It intersects the coordinate axes to form a right-angled triangle. The vertices of this triangle are the origin

(0, 0)

, the x-intercept, and the y-intercept. What must be true about these intercepts for the triangle to exist entirely in the first quadrant?

Hint 2

The Arsenal: You need the equation of the tangent line. Recall that the slope of the tangent to a curve

f (x)

at a point

(x_{1}, y_{1})

is given by the derivative

f^{'} (x_{1})

. Once you have the slope, use the point-slope form

y - y_{1} = m (x - x_{1})

to construct the line equation.

Hint 3

The Execution: After finding the tangent equation

y = (2 + b) x - (1 + b)

, set

x = 0

to find the y-intercept and

y = 0

to find the x-intercept. Use the area formula

A r e a = \frac{1}{2} \times ba se \times h e i g h t = 2

and remember that for the first quadrant, both intercepts must be positive, which imposes a strict constraint on

b

Visualized Solution

Visualizing the Curve and Tangent

Curve: $f (x) = x^{2} + b x - b$
Point of tangency: $P (1, 1)$
The tangent forms a triangle with the coordinate axes in the first quadrant.

Finding the Derivative $f^{'} (x)$

To find the slope of the tangent, we need the derivative of $f (x)$ .
Differentiating $f (x) = x^{2} + b x - b$ with respect to $x$ :
f'(x) = \frac{d}{dx}(x^2 + bx - b) = 2x + b

Slope of Tangent at $P (1, 1)$

The slope $m$ of the tangent at $P (1, 1)$ is the value of $f^{'} (x)$ at $x = 1$ .
m = f'(1) = 2(1) + b
m = 2 + b

Equation of the Tangent Line

Using the point-slope form: $y - y_{1} = m (x - x_{1})$
Substitute $P (1, 1)$ and $m = 2 + b$ :
y - 1 = (2 + b)(x - 1)
y = (2 + b)x - (1 + b)

Finding the $x$ -intercept

To find where the tangent cuts the $x$ -axis, set $y = 0$ :
0 = (2 + b)x - (1 + b)
(2 + b)x = 1 + b
x = \frac{1 + b}{2 + b}

Finding the $y$ -intercept

To find where the tangent cuts the $y$ -axis, set $x = 0$ :
y = (2 + b)(0) - (1 + b)
y = -(1 + b)

First Quadrant Constraints

Since the triangle lies in the first quadrant, both intercepts must be positive:
y\text{-intercept} > 0 \Rightarrow -(1 + b) > 0 \Rightarrow b < -1
x\text{-intercept} > 0 \Rightarrow \frac{1 + b}{2 + b} > 0
Since $1 + b < 0$ , we must have $2 + b < 0 \Rightarrow b < - 2$

Setting up the Area Equation

Area of a right-angled triangle: $A = \frac{1}{2} \times base \times height$
A = \frac{1}{2} \times x_{int} \times y_{int} = 2
\frac{1}{2} \times \left(\frac{1 + b}{2 + b}\right) \times (-(1 + b)) = 2

Simplifying the Equation

Multiply both sides by $2$ :
-\frac{(1 + b)^2}{2 + b} = 4
-(1 + 2b + b^2) = 4(2 + b)
-1 - 2b - b^2 = 8 + 4b

Forming the Quadratic Equation

Rearranging all terms to one side:
b^2 + 4b + 2b + 8 + 1 = 0
b^2 + 6b + 9 = 0

Solving for $b$

Recognize the perfect square:
b^2 + 6b + 9 = (b + 3)^2
(b + 3)^2 = 0 \Rightarrow b + 3 = 0
b = -3

Final Verification and Summary

For $b = - 3$ , the tangent is $y = - x + 2$ .
Intercepts are $(2, 0)$ and $(0, 2)$ .
Area $= \frac{1}{2} \times 2 \times 2 = 2$ .
The correct option is $- 3$ .

00:00 / 00:00

The Sigma Insight: Tangents, Normals and Rate Measure

Solution Diagram

Analyzing the Setup

The curve is defined by the function

f (x) = x^{2} + b x - b

. We are interested in the tangent line at the point

P (1, 1)

For the point

P (1, 1)

to lie on the curve, it must satisfy the equation

f (1) = 1

. Substituting these values:

1 = (1)^{2} + b (1) - b

1 = 1 + b - b

1 = 1

This confirms that the point

P (1, 1)

lies on the curve for any value of

b

The Slope Machine

To find the slope of the tangent, we calculate the derivative of the function

f (x) = x^{2} + b x - b

. Using the power rule, we obtain:

f^{'} (x) = 2 x + b

At the point of tangency

x = 1

, the slope

m

is:

m = f^{'} (1) = 2 (1) + b = 2 + b

Constructing the Tangent

Using the point-slope form

y - y_{1} = m (x - x_{1})

with point

P (1, 1)

and slope

m = 2 + b

, we write:

y - 1 = (2 + b) (x - 1)

Expanding and simplifying this equation:

y = (2 + b) x - (2 + b) + 1

y = (2 + b) x - (1 + b)

The Intercepts and the Constraint

The y-intercept occurs at

x = 0

. Substituting this into the tangent equation:

y_{in t} = - (1 + b)

The x-intercept occurs at

y = 0

. Solving

0 = (2 + b) x - (1 + b)

for

x

x_{in t} = \frac{1 + b}{2 + b}

For the triangle to exist in the first quadrant, both intercepts must be positive. Since

y_{in t} = - (1 + b) > 0

, we must have

b < - 1

. For

x_{in t} = \frac{1 + b}{2 + b} > 0

given

b < - 1

, the denominator must also be negative, leading to the constraint $b < - 2$ .

The Area Equation

The area of the right-angled triangle formed by the axes and the tangent is given by:

A r e a = \frac{1}{2} \times ∣ x_{in t} ∣ \times ∣ y_{in t} ∣ = 2

Substituting our expressions:

\frac{1}{2} \times (\frac{1 + b}{2 + b}) \times (- (1 + b)) = 2

- \frac{( 1 + b ) ^{2}}{2 + b} = 4

Final Calculation

Rearranging the equation to solve for

b

- (1 + b)^{2} = 4 (2 + b)

- (1 + 2 b + b^{2}) = 8 + 4 b

- 1 - 2 b - b^{2} = 8 + 4 b

Combining like terms results in a quadratic equation:

b^{2} + 6 b + 9 = 0

(b + 3)^{2} = 0

Solving for

b

, we find $b = - 3$ . Since this satisfies our constraint

b < - 2

, the final value is $b = - 3$ .

Common Traps & FAQs

Wait, why do we need to check the first quadrant condition? Can't we just solve the area equation?

That is a great question! While the algebra might give you a value for

b

, the geometry of the problem is the master. The area formula

A r e a = \frac{1}{2} \times ba se \times h e i g h t

assumes that the base and height are positive lengths. If the intercepts were negative, the triangle would not be in the first quadrant, and the area calculation would be physically meaningless in this context. Always check your constraints!

Why can't we just use energy conservation or some other physics principle?

I love the enthusiasm for physics! However, this is a pure coordinate geometry and calculus problem. We are dealing with the properties of a curve and its tangent line. The 'energy' here is the mathematical consistency of the tangent equation. Stick to the calculus toolkit—derivatives and line equations—and you will find the path.

What if the quadratic equation had two distinct roots? How would I know which one to pick?

If you had two roots, you would test each one against the constraints we derived from the first-quadrant condition (i.e.,

b < - 2

). If one root satisfies the condition and the other does not, you reject the invalid one. In this specific problem, the quadratic is a perfect square, which simplifies things, but always be ready to filter your solutions!

Animated Solution for Mathematics - Differentiation: The triangle formed by the tangent to the curve $f (x) = x^{2} + b x - b$ at the point $(1, 1)$ and the coordinate axes, lies in the first quadrant. If its area is $2$ , then the value of $b$ is

Select Answer:

Visualized Solution

Visualizing the Curve and Tangent

Finding the Derivative $f^{'} (x)$

Slope of Tangent at $P (1, 1)$

Equation of the Tangent Line

Finding the $x$ -intercept

Finding the $y$ -intercept

First Quadrant Constraints

Setting up the Area Equation

Simplifying the Equation

Forming the Quadratic Equation

Solving for $b$

Final Verification and Summary

The Sigma Insight: Tangents, Normals and Rate Measure

Analyzing the Setup

The Slope Machine

Constructing the Tangent

The Intercepts and the Constraint

The Area Equation

Final Calculation

Similar Questions

If the tangent to the curve $y = x + sin y$ at a point $(a, b)$ is parallel to the line joining $(0, \frac{3}{2})$ and $(\frac{1}{2}, 2)$ , then:

If $∣ f (x_{1}) - f (x_{2}) ∣ < (x_{1} - x_{2})^{2}$ , for all $x_{1}, x_{2} \in R$ . Find the equation of tangent to the curve $y = f (x)$ at the point $(1, 2)$ .

If the curves $y^{2} = 6 x, 9 x^{2} + b y^{2} = 16$ intersect each other at right angles, then the value of $b$ is :

Let the quadratic curve passing through the point $(- 1, 0)$ and touching the line $y = x$ at $(1, 1)$ be $y = f (x)$ . Then the $x$ -intercept of the normal to the curve at the point $(α, α + 1)$ in the first quadrant is

If the line $a x + b y + c = 0$ is a normal to the curve $x y = 1$ , then

If the lines $x + y = a$ and $x - y = b$ touch the curve $y = x^{2} - 3 x + 2$ at the points where the curve intersects the $x$ -axis, then $\frac{a}{b}$ is equal to

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 ________.

If the tangent to the curve $y = x^{3} - x^{2} + x$ at the point $(a, b)$ is also tangent to the curve $y = 5 x^{2} + 2 x - 25$ at the point $(2, - 1)$ , then $∣2 a + 9 b ∣$ is equal to ________

Let $S$ be the set of all values of $x$ for which the tangent to the curve $y = f (x) = x^{3} - x^{2} - 2 x$ at $(x, y)$ is parallel to the line segment joining the points $(1, f (1))$ and $(- 1, f (- 1))$ , then $S$ is equal to :

Select Answer:

Visualized Solution

Visualizing the Curve and Tangent

Finding the Derivative .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(x)

Slope of Tangent at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(1,1)

Equation of the Tangent Line

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

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

First Quadrant Constraints

Setting up the Area Equation

Simplifying the Equation

Forming the Quadratic Equation

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

Final Verification and Summary

The Sigma Insight: Tangents, Normals and Rate Measure

Analyzing the Setup

The Slope Machine

Constructing the Tangent

The Intercepts and the Constraint

The Area Equation

Final Calculation

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∣f(x1​)−f(x2​)∣<(x1​−x2​)2, for all x1​,x2​∈R. Find the equation of tangent to the curve y=f(x) at the point (1,2).

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the curves y2=6x,9x2+by2=16 intersect each other at right angles, then the value of b is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the line ax+by+c=0 is a normal to the curve xy=1, then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the lines x+y=a and x−y=b touch the curve y=x2−3x+2 at the points where the curve intersects the x-axis, then ba​ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the tangent to the curve y=x3−x2+x at the point (a,b) is also tangent to the curve y=5x2+2x−25 at the point (2,−1), then ∣2a+9b∣ is equal to ________

Finding the Derivative $f^{'} (x)$

Slope of Tangent at $P (1, 1)$

Finding the $x$ -intercept

Finding the $y$ -intercept

Solving for $b$

If the tangent to the curve $y = x + sin y$ at a point $(a, b)$ is parallel to the line joining $(0, \frac{3}{2})$ and $(\frac{1}{2}, 2)$ , then:

If $∣ f (x_{1}) - f (x_{2}) ∣ < (x_{1} - x_{2})^{2}$ , for all $x_{1}, x_{2} \in R$ . Find the equation of tangent to the curve $y = f (x)$ at the point $(1, 2)$ .

If the curves $y^{2} = 6 x, 9 x^{2} + b y^{2} = 16$ intersect each other at right angles, then the value of $b$ is :

If the line $a x + b y + c = 0$ is a normal to the curve $x y = 1$ , then

If the lines $x + y = a$ and $x - y = b$ touch the curve $y = x^{2} - 3 x + 2$ at the points where the curve intersects the $x$ -axis, then $\frac{a}{b}$ is equal to

If the tangent to the curve $y = x^{3} - x^{2} + x$ at the point $(a, b)$ is also tangent to the curve $y = 5 x^{2} + 2 x - 25$ at the point $(2, - 1)$ , then $∣2 a + 9 b ∣$ is equal to ________