Question
The eccentricity of the hyperbola with asymptotes 3x + 4y = 2 and 4x – 3y = 2 is
- 3
- 2
- 4
Hint:
The eccentricity of a hyperbola is always greater than 1. i.e. e > 1. The eccentricity of a hyperbola can be taken as the ratio of the distance of the point on the hyperbole, from the focus, and its distance from the directrix.
Eccentricity = Distance from Focus/Distance from Directrix
The correct answer is: ![square root of 2](data:image/png;base64,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)
![square root of 2](data:image/png;base64,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)
We have been given two asymptotes i.e. 3x + 4y = 2 and 4x – 3y = 2
First find out their slopes
![m 1 space equals negative space 3 over 4 space a n d space m 2 space equals space 4 over 3](data:image/png;base64,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)
From above we can see that m1 and m2 are negative inverse of each other that means both are perpendicular to each other.
It is a rectangular hyperbola and it has fixed eccentricity.
Thus eccentricity = ![square root of 2](data:image/png;base64,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)
Related Questions to study
The number of ways that '6' rings can be worn on the 4 - fingers of one hand is
It is important to note that we have used a basic fundamental principle of counting to find the total ways. Also, it is important to notice that each ring has 4 ways as it has not been given that each finger must have at least one ring. So, there can be 6 rings in a finger alone and remaining all the fingers empty.
The number of ways that '6' rings can be worn on the 4 - fingers of one hand is
It is important to note that we have used a basic fundamental principle of counting to find the total ways. Also, it is important to notice that each ring has 4 ways as it has not been given that each finger must have at least one ring. So, there can be 6 rings in a finger alone and remaining all the fingers empty.