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An asymptote is defined as a line that is straight with respect to a curve in a way that it meets the curve of the plane at infinity. It can also be described as a line that is drawn at the minimum distance which is parallel to the tangent of a curve, in a way that it does not touch or cut the curve. Asymptote formula is defined generally for a hyperbola. It is shown as an equation of a line. It is generally noticed that one curve is of another curve’s curvilinear asymptote, which is the opposite of a linear asymptote, that is when the distance between both of the curves approaches zero as they approach infinity. The term asymptote however is itself fixed for linear asymptotes.

Table of Content

What is the Asymptote Formula?
Horizontal Asymptote
Vertical Asymptote
Oblique Asymptote
Equation of Asymptotes
Asymptotes of a Hyperbola
Things to Remember
Sample Questions

Key Words: Asymptote, Formula, Curve, Straight, Line, Infinity, Constant.

What is the Asymptote Formula?

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An asymptote that is of the curve y = f(x) or is in an implicit form: f(x,y) = 0 is described as a straight line in a way that the distance between the straight line and the curve gets closer to zero when the points on the curve get closer to infinity.

The procedure chosen to find the horizontal asymptote may change according to the degrees of the polynomials in the denominator and the numerator of the function. If both of the polynomials’ degrees are found to be the same, then we shall divide the coefficients of the degree terms that are the largest.

Asymptote of a curve

Asymptote of a curve

The three kinds of asymptotes are as follows:

Vertical Asymptotes
Horizontal Asymptotes
Oblique Asymptotes

Horizontal Asymptote

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In Horizontal Asymptotes, The curve will approach a constant value b, when x moves towards -∞ or ∞.

Horizontal Curve

Horizontal asymptote

Vertical Asymptote

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In the Vertical Asymptote, the curve moves towards the direction of -∞ or ∞ when x tends to approach a constant value c from left or right.

Vertical Asymptote

Vertical asymptote

Oblique Asymptote

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In the Oblique Asymptote, the curve moves towards the direction of the line y = mx + b, when x moves towards either -∞ or ∞.

Oblique Asymptote

Oblique asymptote

It is worth noting that m here is not zero as the Asymptote is Horizontal.

Equation of Asymptotes

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For Vertical Asymptote

It consists of a straight line with the equation x = a for the graph of function y = f(x), however, it must satisfy at least one of the conditions given below:

Lim x→a+0 f(x)=±∞

or Lim x→a−0 f(x)=±∞

If not, then at least one of the limits being one-sided at the point x = a must be equal to infinity.

For Oblique Asymptote.

The graph of function y=f(x) for the equation of straight-line is y=kx+b for the limit x→+∞

But only if the two limits given below are finite.

Lim x→+∞ f(x)/x=k and Lim x→+∞ [f(x)−kx]=b

For Horizontal Asymptote

The graph of function y=f(x) has the straight line with the equation is y=b, which is known as the asymptote of a function x→+∞, but only when the given limit is finite.

Lim x→+∞ f(x) = b

Asymptotes of a Hyperbola

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The asymptotes of the hyperbola are shown as a pair of straight lines. The asymptotes of a hyperbola that have an equation as x2/a2 - y2/b2 = 0 is shown by the given following formula:

The Equation of Asymptotes is b/a.x, and y = -b/a.x

Equation of Pair of Asymptotes is x2/a2 - y2/b2 = 0

Asymptote of Hyperbola

Asymptote of hyperbola

Things To Remember

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An asymptote of a curve, y = f(x) is described as a straight line in a way that the distance between the straight line and the curve gets closer to zero when the points on the curve get closer to infinity.
It can also be described as a line that is drawn at the minimum distance which is parallel to the tangent of a curve, in a way that it does not touch or cut the curve.
If the hyperbola’s centre is at the origin, then the pair of asymptotes is y = ±(b/a)x
That means, y = (b/a)x and y = -(b/a)x.
The Asymptotes are of three kinds, namely: Vertical, Horizontal and Oblique.

Sample Questions

Ques. Find the equation of a pair of asymptotes of a hyperbola x2/16 - y2/25 = 1. [3 marks]

Ans. Given the equation of the hyperbola is x2/16 - y2/25 = 1

For a hyperbola having an equation x2/a2 - y2/b2 = 1 the equation of its pair of asymptotes is bx/a and -bx/a

We know that b = 5 and a = 4.

Therefore, the equation of the pair of asymptotes is y =-4x/5 and y = 4x/5.

Therefore, the Equation of the pair of asymptotes is 5y-4x = 0 and 5y +4x =0

Ques. Find the equations of the asymptotes of the hyperbola x2/49 - y2/36 = 1. [3 marks]

Ans. The given equation of the hyperbola is x2/49 - y2/36 = 1

→ x2/72 - y2/62 = 1

Upon comparing the above equation with the standard equation of a hyperbola x2/a2 - y2/b2 = 1

We find that b = 6 and a = 7.

And the equation of the asymptotes are found to be y = -b/a.x and y = b/a.x.

y = 6x/7 and y = -6x/7

7y = 6x and 7y = -6x

7y- 6x = 0 and 6x + 7y = 0

Hence the equations of the asymptotes are 7y- 6x = 0 and 6x + 7y = 0

Ques. Find the Asymptotes parallel to y-axis for curve a2x2 − b2y2 = 1. [5 marks]

Ans. The Given curve is a2/x2 − b2/y2 = 1

a2y2 - b2x2 = x2y2

a2y2 - x2y2 = b2x2

y2(a2 - x2 ) = b2x2

y2 = b2x2/a2 −x2

y = ±bx (a2−x2)

An asymptote is called a horizontal asymptote when it is parallel to the x-axis If the degrees of the denominator and the numerator are identical, then the horizontal asymptote is equal to the coefficient of the largest exponent of the numerator that is divided by the denominator’s lead coefficient y = ±b.

Ques. Given the function g(x) = x/x2 + 2, determine its horizontal asymptotes. [ 2marks]

Ans. In this function, we observe that the numerator has x’s degree less than that of the degree of x in the denominator.

This means that the horizontal asymptote is located at y = 0

Ques. Given the function f(x) = x2 + 2/x + 1, find its horizontal asymptotes. [3 marks]

Ans. As we know, if the degree of the numerator is greater than the degree of the denominator, then the function does not contain a horizontal asymptote.

Now, we shall compare the above equation to this statement.

Upon closer inspection, we find that the equation’s numerator has a degree of 2 while the denominator only has a degree of one.

So We conclude the fact that this function does not have a horizontal asymptote:

Ques. Find the horizontal asymptote of the function f(x) = 2x + 1/ 3x - 5. [3 marks]

Ans. Given: Function is f(x)= 2x + 1/ 3x - 5.

We observe that the degree of x in the numerator is the same as the degree of x in the denominator.

So now we shall Divide the leading coefficients of both the numerator and the denominator.

So, the line y = 2/3 is the horizontal asymptote.

Ques. Find the vertical Asymptote of f(x)= 3x2 + 6x + 5/x2 - 3x + 2. [3 marks]

Ans. Let the denominator be equal to 0 and then solve,

x2 - 3x + 2 = 0

This equation can be solved if we factor the trinomial and set the factors to be equal to 0.

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

x - 2 = 0 and x - 1 = 0

x = 2 and x = 1.

So, The vertical asymptotes are at x = 2 and x = 1.

Ques. What is the method of identifying a vertical graph? [4 marks]

Ans. The method of identifying a vertical graph is as follows:

Firstly we have to Factorise the denominator and numerator.
Find out any restrictions in the function’s domain.
Now, cancel the common factors between the numerator and the denominator for Reducing the expression.
Keep Note of the values that make the denominator zero in after canceling out. These are points the vertical asymptotes shall be found.
Asymptotes do not occur in the restrictions of the domain. These are known as removable discontinuities.

Ques. Find the Equation of hyperbola whose vertices are (± 5, 0) and its asymptotes are 3x ± 5y = 0. [5 marks]

Ans. As we know, The equation of the hyperbola is x2/a2−y2/b2=1.

The vertices here are (± a, 0).

The asymptotes are shown as the straight lines y = -(b/a)x and y = (b/a)x.

The given Asymptotes are: 3x ± 5y = 0 and Vertices are (± 5, 0).

So, 5y = -3x and 5y = 3x

y=−(3/5)x and y=(3/5)x

∴ b = 3 and a = 5

So, the Equation of hyperbola is x2/25 − y2/9 = 1

9x2 – 25y2 = 225

Ques. What are the asymptotes of the hyperbola 9x2 - 16y2 = 144? [3 marks]

Ans. The given equation of the hyperbola is 9x2 - 16y2 = 144

This above equation can be written as x2/16 − y2/9 = 1

Now we compare the above equation with x2/a2 − y2/b2 =1

And we get ⇒ b = 3 and a = 4.

So, the asymptotes are y = ±3/4x

Related links

Equation for hyperbola	Difference between parabola and hyperbola	Limit and continuity
Calculus formula	Differentiation and integration formula	Introduction to 3-dimensional geometry
Parabola, hyperbola, ellipse formula	Circle	Tangents and normals
Formation of differential equation	Conditional probability formula	Conic section

Vertical, horizontal Asymptote: Formula, Solved questions

What is the Asymptote Formula?

Horizontal Asymptote

Vertical Asymptote

Oblique Asymptote

Equation of Asymptotes

Asymptotes of a Hyperbola

Things To Remember

Sample Questions

CBSE CLASS XII Related Questions

1.
Find: \[ \int \frac{x^2}{(x^2-1)(x^2+4)}\,dx \]

2.
Find:

The shortest distance between the lines: \[ \vec{r}=(4+\lambda)\hat{i}+(2\lambda-1)\hat{j}-3\lambda\hat{k} \] and \[ \vec{r}=(1+2\mu)\hat{i}+(4\mu-1)\hat{j}+(2-5\mu)\hat{k} \]

4.
Find:

If the function \[ f(x)= \begin{cases} \frac{\sin x}{x}+\cos x, & x\neq0\\ k, & x=0 \end{cases} \] is continuous at \(x=0\), then find the value of \(k\).

5.
Evaluate:
\[ \int_{0}^{1} \frac{x \tan^{-1}x}{(1+x^2)^{3/2}}\,dx \]

6.
A carpenter needs to design a wooden box in the shape of a cuboid such that the sum of its length and breadth is 3 cm more than its height. Twice of its length, thrice of its breadth and its height add up to 10 cm. Its breadth added to 7 times its height is 1 cm less than 3 times its length.

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Vertical, horizontal Asymptote: Formula, Solved questions

What is the Asymptote Formula?

Horizontal Asymptote

Vertical Asymptote

Oblique Asymptote

Equation of Asymptotes

Asymptotes of a Hyperbola

Things To Remember

Sample Questions

CBSE CLASS XII Related Questions

1.Find: \[ \int \frac{x^2}{(x^2-1)(x^2+4)}\,dx \]

2. Find: The shortest distance between the lines: \[ \vec{r}=(4+\lambda)\hat{i}+(2\lambda-1)\hat{j}-3\lambda\hat{k} \] and \[ \vec{r}=(1+2\mu)\hat{i}+(4\mu-1)\hat{j}+(2-5\mu)\hat{k} \]

4. Find: If the function \[ f(x)= \begin{cases} \frac{\sin x}{x}+\cos x, & x\neq0\\ k, & x=0 \end{cases} \] is continuous at \(x=0\), then find the value of \(k\).

5.Evaluate:\[ \int_{0}^{1} \frac{x \tan^{-1}x}{(1+x^2)^{3/2}}\,dx \]

6.A carpenter needs to design a wooden box in the shape of a cuboid such that the sum of its length and breadth is 3 cm more than its height. Twice of its length, thrice of its breadth and its height add up to 10 cm. Its breadth added to 7 times its height is 1 cm less than 3 times its length.

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SUBSCRIBE TO OUR NEWS LETTER

1.
Find: \[ \int \frac{x^2}{(x^2-1)(x^2+4)}\,dx \]

2.
Find:

The shortest distance between the lines: \[ \vec{r}=(4+\lambda)\hat{i}+(2\lambda-1)\hat{j}-3\lambda\hat{k} \] and \[ \vec{r}=(1+2\mu)\hat{i}+(4\mu-1)\hat{j}+(2-5\mu)\hat{k} \]

4.
Find:

If the function \[ f(x)= \begin{cases} \frac{\sin x}{x}+\cos x, & x\neq0\\ k, & x=0 \end{cases} \] is continuous at \(x=0\), then find the value of \(k\).

5.
Evaluate:
\[ \int_{0}^{1} \frac{x \tan^{-1}x}{(1+x^2)^{3/2}}\,dx \]

6.
A carpenter needs to design a wooden box in the shape of a cuboid such that the sum of its length and breadth is 3 cm more than its height. Twice of its length, thrice of its breadth and its height add up to 10 cm. Its breadth added to 7 times its height is 1 cm less than 3 times its length.