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Animated Solution for Mathematics - Differentiation: Angle between the tangents to the curve $y = x^{2} - 5 x + 6$ at the points $(2, 0)$ and $(3, 0)$ is

Q: Why do we differentiate to find the slope? Can't we just use the slope formula $\frac{y_2 - y_1}{x_2 - x_1}$?

The slope formula $\frac{y_2 - y_1}{x_2 - x_1}$ is for the slope of a secant line connecting two points. But here, we need the slope of a tangent line at a single point. Differentiation is the tool that gives us the instantaneous rate of change, which is exactly the slope of the tangent at that specific point.

Q: Is the angle between tangents always $\frac{\pi}{2}$ for any parabola?

Absolutely not! It only happens to be $\frac{\pi}{2}$ here because of the specific points chosen. For this parabola, the tangents at the roots happen to be perpendicular. For other points, the angle will be different. Always calculate the slopes first!

Select Answer:

Visualized Solution

Visualizing the Curve

Given curve: $y = x^{2} - 5 x + 6$
Points of interest: $P_{1} (2, 0)$ and $P_{2} (3, 0)$
Objective: Find the angle $θ$ between the tangents at these points.

The Slope Tool: Differentiation

To find the angle between tangents, we first need their slopes.
The slope of a curve at any point is given by its derivative, $\frac{d y}{d x}$ .

Setting up the Derivative

We need to differentiate the equation of the curve with respect to $x$ .
$\frac{d y}{d x} = \frac{d}{d x} (x^{2} - 5 x + 6)$

Calculating $\frac{d y}{d x}$

$\frac{d}{d x} (x^{2}) = 2 x$
$\frac{d}{d x} (- 5 x) = - 5$
$\frac{d y}{d x} = 2 x - 5$

Slope at $P_{1} (2, 0)$ - Setup

At point $P_{1} (2, 0)$ , the $x$ -coordinate is $2$ .
Substitute $x = 2$ into $\frac{d y}{d x}$ to find the first slope, $m_{1}$ .
$m_{1} = 2 (2) - 5$

Evaluating Slope $m_{1}$

$m_{1} = 4 - 5$
$m_{1} = - 1$
The slope of the first tangent is $- 1$ .

Slope at $P_{2} (3, 0)$ - Setup

At point $P_{2} (3, 0)$ , the $x$ -coordinate is $3$ .
Substitute $x = 3$ into $\frac{d y}{d x}$ to find the second slope, $m_{2}$ .
$m_{2} = 2 (3) - 5$

Evaluating Slope $m_{2}$

$m_{2} = 6 - 5$
$m_{2} = 1$
The slope of the second tangent is $1$ .

Analyzing the Slopes

We have found the slopes of both tangents:
$m_{1} = - 1$
$m_{2} = 1$
Notice the relationship between these two values.

Product of Slopes

Let's calculate the product of the two slopes, $m_{1} \cdot m_{2}$ .
$m_{1} \cdot m_{2} = (- 1) \cdot (1)$
$m_{1} \cdot m_{2} = - 1$

Final Conclusion

Since $m_{1} \cdot m_{2} = - 1$ , the tangents are perpendicular.
The angle between them is $9 0^{\circ}$ or $\frac{π}{2}$ radians.
Correct Option: $\frac{π}{2}$

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The Sigma Insight: Tangents, Normals and Rate Measure

Solution Diagram

Analyzing the Setup

Imagine you are standing on the graph of the parabola

y = x^{2} - 5 x + 6

. This curve dips down and then rises, crossing the x-axis at two distinct points:

(2, 0)

and

(3, 0)

Our mission is to determine the angle between the tangents drawn to the curve at these two specific locations. A tangent line represents the instantaneous direction of the curve at a given point.

The Slope-Generating Machine

To find the angle between two lines, we must first determine their slopes. In calculus, the slope of a curve at any point is given by the derivative,

\frac{d y}{d x}

We differentiate the curve

y = x^{2} - 5 x + 6

using the power rule:

\frac{d y}{d x} = 2 x - 5

This expression,

2 x - 5

, acts as a slope-generating machine. By substituting an x-coordinate into this derivative, we obtain the slope of the tangent at that specific point.

Calculating the Slopes

First, we evaluate the slope at the point

P_{1} (2, 0)

by substituting

x = 2

m_{1} = 2 (2) - 5 = 4 - 5 = - 1

The slope of the first tangent is

- 1

, indicating that the line is tilting downwards.

Next, we evaluate the slope at the point

P_{2} (3, 0)

by substituting

x = 3

m_{2} = 2 (3) - 5 = 6 - 5 = 1

The slope of the second tangent is

1

, indicating that the line is tilting upwards at a

4 5^{\circ}

angle.

The Geometric Revelation

We have identified the two slopes as

m_{1} = - 1

and

m_{2} = 1

. We now examine the product of these slopes:

m_{1} \cdot m_{2} = (- 1) \cdot (1) = - 1

In coordinate geometry, if the product of the slopes of two lines is

- 1

, the lines are perfectly perpendicular to each other. They intersect at a right angle.

Therefore, the angle between the tangents at

(2, 0)

and

(3, 0)

is $9 0^{\circ}$ (or $\frac{π}{2}$ radians). This result highlights the inherent symmetry of the parabola and the elegance of its geometric properties.

Common Traps & FAQs

Why do we differentiate to find the slope? Can't we just use the slope formula

\frac{y _{2} - y _{1}}{x _{2} - x _{1}}

The slope formula

\frac{y _{2} - y _{1}}{x _{2} - x _{1}}

is for the slope of a secant line connecting two points. But here, we need the slope of a tangent line at a single point. Differentiation is the tool that gives us the instantaneous rate of change, which is exactly the slope of the tangent at that specific point.

Wait, why is the product of slopes

m_{1} \cdot m_{2} = - 1

so important?

That is the classic condition for orthogonality! In coordinate geometry, if two lines have slopes

m_{1}

and

m_{2}

, they are perpendicular if and only if

m_{1} \cdot m_{2} = - 1

. It is a beautiful, elegant shortcut that saves us from using the more complex tangent angle formula

tan θ = ∣ \frac{m _{1} - m _{2}}{1 + m _{1} m _{2}} ∣

Is the angle between tangents always

\frac{π}{2}

for any parabola?

Absolutely not! It only happens to be

\frac{π}{2}

here because of the specific points chosen. For this parabola, the tangents at the roots happen to be perpendicular. For other points, the angle will be different. Always calculate the slopes first!

Animated Solution for Mathematics - Differentiation: Angle between the tangents to the curve $y = x^{2} - 5 x + 6$ at the points $(2, 0)$ and $(3, 0)$ is

Select Answer:

Visualized Solution

Visualizing the Curve

The Slope Tool: Differentiation

Setting up the Derivative

Calculating $\frac{d y}{d x}$

Slope at $P_{1} (2, 0)$ - Setup

Evaluating Slope $m_{1}$

Slope at $P_{2} (3, 0)$ - Setup

Evaluating Slope $m_{2}$

Analyzing the Slopes

Product of Slopes

Final Conclusion

The Sigma Insight: Tangents, Normals and Rate Measure

Analyzing the Setup

The Slope-Generating Machine

Calculating the Slopes

The Geometric Revelation

Similar Questions

If $θ$ denotes the acute angle between the curves, $y = 10 - x^{2}$ and $y = 2 + x^{2}$ at a point of their intersection, then $∣ tan θ ∣$ is equal to :

The tangents to the curve $y = (x - 2)^{2} - 1$ at its points of intersection with the line $x - y = 3$ , intersect at the point :

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:

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

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 :

Let $θ$ be the acute angle between the tangents to the ellipse $\frac{x ^{2}}{9} + \frac{y ^{2}}{1} = 1$ and the circle $x^{2} + y^{2} = 3$ at their point of intersection in the first quadrant. Then $tan θ$ is equal to :

An angle of intersection of the curves, $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ and $x^{2} + y^{2} = ab, a > b$ , is :

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 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 ________

Consider: $f (x) = tan^{- 1} (\frac{1 + s i n x}{1 - s i n x}), x \in (0, \frac{π}{2})$ . A normal to $y = f (x)$ at $x = \frac{π}{6}$ also passes through the point

Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Angle between the tangents to the curve y=x2−5x+6 at the points (2,0) and (3,0) is

Select Answer:

Visualized Solution

Visualizing the Curve

The Slope Tool: Differentiation

Setting up the Derivative

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

Slope at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P1​(2,0) - Setup

Evaluating Slope .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m1​

Slope at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P2​(3,0) - Setup

Evaluating Slope .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m2​

Analyzing the Slopes

Product of Slopes

Final Conclusion

The Sigma Insight: Tangents, Normals and Rate Measure

Analyzing the Setup

The Slope-Generating Machine

Calculating the Slopes

The Geometric Revelation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If θ denotes the acute angle between the curves, y=10−x2 and y=2+x2 at a point of their intersection, then ∣tanθ∣ is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The tangents to the curve y=(x−2)2−1 at its points of intersection with the line x−y=3, intersect at the point :

.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:

.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=x2−3x​,x∈R,x=±3​, at a point (α,β)=(0,0) on it is parallel to the line 2x+6y−11=0, then :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } An angle of intersection of the curves, a2x2​+b2y2​=1 and x2+y2=ab,a>b, is :

.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 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 ________

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Consider: f(x)=tan−1(1−sinx1+sinx​​),x∈(0,2π​). A normal to y=f(x) at x=6π​ also passes through the point

Animated Solution for Mathematics - Differentiation: Angle between the tangents to the curve $y = x^{2} - 5 x + 6$ at the points $(2, 0)$ and $(3, 0)$ is

Calculating $\frac{d y}{d x}$

Slope at $P_{1} (2, 0)$ - Setup

Evaluating Slope $m_{1}$

Slope at $P_{2} (3, 0)$ - Setup

Evaluating Slope $m_{2}$

If $θ$ denotes the acute angle between the curves, $y = 10 - x^{2}$ and $y = 2 + x^{2}$ at a point of their intersection, then $∣ tan θ ∣$ is equal to :

The tangents to the curve $y = (x - 2)^{2} - 1$ at its points of intersection with the line $x - y = 3$ , intersect at the point :

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:

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 :

An angle of intersection of the curves, $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ and $x^{2} + y^{2} = ab, a > b$ , is :

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 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 ________

Consider: $f (x) = tan^{- 1} (\frac{1 + s i n x}{1 - s i n x}), x \in (0, \frac{π}{2})$ . A normal to $y = f (x)$ at $x = \frac{π}{6}$ also passes through the point