## IIT JEE preparation tips of Asymptote Of Hyperbola

IIT JEE preparation tips : An hyperbole is an open curve with two branches obtained by cutting a straight cone with a plain that is not necessarily parallel to the symmetry axis, and with a smaller angle to the revolution axis that the angle of the generatrix. In analytical geometry, an hyperbole is the geometric place of the points in a plain, given that the absolute value of the difference of their distances to two fixated points, named focus, is equal to the distance between the vertex, which is a positive constant.

The asymptote of an hyperbole (A1 and A2) are the two straight lines that approach closer and closer to the hyperbole but they can never reach it and cut through it. It is known that in infinity the asymptotes will be at a distance equal to zero from the hyperbole.

The equations of the asymptotes can be obtained by calculating the real semiaxis(a) and the imaginary semiaxis (b).

** Read More IIT JEE preparation tips : Location Of Roots**

## Example 1.

I n the image you can see an hyperbole, with known semiaxis, being the real semiaxis 2 * a*=4 cm and 2

*=8 cm. These two asymptotes are defined by the following formulas:*

*b*

**Equations for hyperboles with non-origin center**

The algebraic procedure for the deduction of the equations of the hyperbole which center is in any point in space but the point of origin is similar to the procedure used to find the equations of hyperboles whose center is in the coordinates origin and we will leave it as an exercise for students. The corresponding equations to the two possible scenarios in the real axis (horizontal or vertical) are as follows:

### Real Horizontal axis (parallel to X axis)

[(x – h)^{2}/a^{2}] – [(y – k)^{2}/b^{2}] = 1

### Real vertical axis (parallel to Y axis)

[(y – k)^{2}/a^{2}] – [(x – h)^{2}/b^{2}] = 1

In mathematics, the graph of a straight function that continuously approaches the graph of another function is called asymptote, as long as the distance between the two tends to be zero (0), while they stretch towards infinity. Or that both of them present an asymptotic behavior. Generally speaking, all the functions that are rational tend to have asymptotic behavior.

In much the same way, it is possible to demonstrate that the equations of the asymptotes for the two shown cases are as follows:

Real horizontal axis (parallel to X axis)

y – k = ± [b (x – h)/a]

Real vertical axis (parallel to Y axis)

y – k = ± [a (x – h)/b]

** Read More IIT JEE preparation tips : Calculus**

## Example 2.

The center of an hyperbole is in (−3, 2), it’s focal distance is 10 units and one of the vertex is the point P(1, 2). Find it’s equation and determine the coordinates of the focus and the extremes of the imaginary axis, and also the equations of their asymptotes.

In the graph shown in the figure we can visualize that the distance CV2 is equal to 4 (a = 4), we can also see that the focal distance is equal to 10, this is 2c =10, which means c = 5. From the relationship between a,b and c we can determine the value of b:

c^{2} = a^{2} + b^{2 }→ b^{2 }= c^{2} – a^{2} → b^{2 }= (5)^{2} – (4)^{2}

b^{2 }= 25 – 16 → b^{2 }= 9 → ( b^{2 })^{1/2 }= (9 )^{1/2}

**b = 3**

b^{2 }= c^{2} – a^{2 }→ b^{2 }= (5)^{2} – (3)^{2 }→ b^{2 }= 25 – 9

b^{2 }= 16 → ( b^{2 })^{1/2 }= (16 )^{1/2}

**b = 4**

As the real axis must cut through the center, the vertex and the focus, you can deduce that said axis is horizontal by reading the graph, that is why the equation is:

[(x – h)^{2}/a^{2}] – [(y – k)^{2}/b^{2}] = 1

Substituting the values for (-3,2), we now have:

[(x – (-3))^{2}/(4)^{2}] – [(y – 2)^{2}/(3)^{2}] = 1

[(x + 3)^{2}/(4)^{2}] – [(y – 2)^{2}/(3)^{2}] = 1

**[(x + 3)**^{2}**/16] – [(y – 2)**^{2}**/9] = 1**

…which is the equation we were looking for.

For the focus,while watching the figure, if you use the value of c to move forward towards the left side and towards the right side respectively, you obtain the points F1(−8, 2) and F2(2, 2).

For the extremes of the imaginary axis we now use the value of b, moving forward from the center, perpendicularly to the real axis, in this case towards the bottom and towards the top, obtaining the points B1(−3, −1) and B2(−3, 5).

Finally, the equations for the asymptotes are:

y – k = ± [b (x – h)/a]

Using the values for this exercise we obtain:

y – 2 = ± [3 (x – (-3))/4]

y – 2 = ± [3 (x + 3)/4]

For the plus sing we obtain:

y – 2 = 3 (x + 3)/4

**4y – 8 = 3x + 9**

For the minus sing, we get:

y – 2 = – 3 (x + 3)/4

**4y – 8 = 9 – 3x**

**Read More IIT JEE preparation tips : Tips & Tricks For IIT JEE**

## Example 3.

Find the coordinates of the vertex and the focus and also the asymptote equations for the hyperbole with formula 9x² − 16y² = 144.

*Resolution.*

(9x²)/144 − (16y²)/144 = 144/144

(x²)/16 − (y²)/9 = 1

Since (x²)/a^{2} − (y²)/b^{2} = 1

Then: a = 4 and b = 3.

So, the coordinates of the vertex must be **A = (4,0)** and **A’ = (-4,0)**; also, the coordinates of the focus are **F = (5,0)** and **F’=(-5,0)**. It is also fairly simple to deduce the asymptote equations, which in this case are **y = 3x/4** and **y = -3x/4**.

## Example 4.

The non focal axis of an hyperbole is 8 and the equation of the asymptotes are y = 2x/3 and y=-2x/3. Calculate the equation of the hyperbole, it’s axis, focus and vertex.

2b = 8 → b=4

2/3 = 4/a → a=6

Because c^{2} = a^{2} + b^{2},then:

c = (36 + 16)^{1/2} → c = 2(13)^{1/2}

Since (x²)/a^{2} − (y²)/b^{2} = 1

(x^{2}) / 36 – (y^{2})/16 = 1

So, the coordinates of the vertex must be **A = (6,0)** and **A’ = (-6,0)**; also, the coordinates of the focus are **F = [****2(13)**^{1/2}**,0]** and **F’=[-****2(13)**^{1/2}**,0]**.

If you want learn more about Hyperbola Asymptotes Practice questions from Co- Ordinate Geometry by G. Tewani.

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