
What is the eccentricity of a rectangular hyperbola?
(a). \[\sqrt 2 \]
(b). \[\sqrt 3 \]
(c). \[\sqrt 5 \]
(d). \[\sqrt 6 \]
Answer
503.7k+ views
Hint: When the transverse axis of a hyperbola is equal to its conjugate axis then the hyperbola is called a rectangular hyperbola. The eccentricity e of the hyperbola is given by \[e = \dfrac{{\sqrt {{a^2} + {b^2}} }}{a}\]. Use this formula to calculate the eccentricity.
Complete step-by-step solution -
A hyperbola is a curve generated by a point moving such that the difference of the distances from two fixed points is constant. Its equation is given by:
\[\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1\]
If the transverse axis of the hyperbola, represented by a, is equal to its conjugate axis, represented by b, then the hyperbola is called a rectangular hyperbola. The asymptotes are x = y and x = - y. Hence, we have as follows:
\[a = b............(1)\]
Eccentricity is a measure of how nearly circular the curve is. It is defined as the ratio of the distance from the center to the focus and the distance from the center to the vertex.
The eccentricity of a hyperbola is always greater than 1. The eccentricity of a hyperbola with the equation \[\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1\] is given as follows:
\[e = \dfrac{{\sqrt {{a^2} + {b^2}} }}{a}\]
Substituting equation (1) in the above equation, we have:
\[e = \dfrac{{\sqrt {{a^2} + {a^2}} }}{a}\]
Simplifying, we have:
\[e = \dfrac{{\sqrt {2{a^2}} }}{a}\]
\[e = \dfrac{{\sqrt 2 a}}{a}\]
Canceling the common terms in the numerator and denominator, we have:
\[e = \sqrt 2 \]
The eccentricity of a rectangular hyperbola is \[\sqrt 2 \].
Hence, the correct answer is option (a).
Note: You need to know the concepts of the eccentricity of the hyperbola to be able to solve this question. The formula for the eccentricity of a hyperbola is \[e = \dfrac{{\sqrt {{a^2} + {b^2}} }}{a}\]. You can also find the focus first and then find the ratio to get the eccentricity.
Complete step-by-step solution -
A hyperbola is a curve generated by a point moving such that the difference of the distances from two fixed points is constant. Its equation is given by:
\[\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1\]

If the transverse axis of the hyperbola, represented by a, is equal to its conjugate axis, represented by b, then the hyperbola is called a rectangular hyperbola. The asymptotes are x = y and x = - y. Hence, we have as follows:
\[a = b............(1)\]
Eccentricity is a measure of how nearly circular the curve is. It is defined as the ratio of the distance from the center to the focus and the distance from the center to the vertex.
The eccentricity of a hyperbola is always greater than 1. The eccentricity of a hyperbola with the equation \[\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1\] is given as follows:
\[e = \dfrac{{\sqrt {{a^2} + {b^2}} }}{a}\]
Substituting equation (1) in the above equation, we have:
\[e = \dfrac{{\sqrt {{a^2} + {a^2}} }}{a}\]
Simplifying, we have:
\[e = \dfrac{{\sqrt {2{a^2}} }}{a}\]
\[e = \dfrac{{\sqrt 2 a}}{a}\]
Canceling the common terms in the numerator and denominator, we have:
\[e = \sqrt 2 \]
The eccentricity of a rectangular hyperbola is \[\sqrt 2 \].
Hence, the correct answer is option (a).
Note: You need to know the concepts of the eccentricity of the hyperbola to be able to solve this question. The formula for the eccentricity of a hyperbola is \[e = \dfrac{{\sqrt {{a^2} + {b^2}} }}{a}\]. You can also find the focus first and then find the ratio to get the eccentricity.
Recently Updated Pages
How do you factor x2 + x 20 0 class 9 maths CBSE

How do you solve y6x and 2x+3y20 using substitutio class 9 maths CBSE

Chipko movement originated in Gopeshwar in A 1953 B class 9 biology CBSE

The adjacent sides in the parallelogram are supplementary class 9 maths CBSE

The compound used in plastic industry is A Vinyl acetate class 9 chemistry CBSE

How do you solve for y in 2left y dfrac12 right 4left class 9 maths CBSE

Trending doubts
According to Bernoullis equation the expression which class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

Differentiate between calcination and roasting class 11 chemistry CBSE

A solution of a substance X is used for white washing class 11 chemistry CBSE

What is spore formation class 11 biology CBSE

10 examples of friction in our daily life
