Question

Find the equation of the hyperbola with centre at origin, transverse axis along x-

axis, eccentricity √5 and sum of whose semi-axis is 9.​

Answer 1

Final Answer:

The equation of the hyperbola with center at origin, the transverse axis along x-axis, and the eccentricity $\sqrt{5}$  such that the sum of the semi-axes is $9$ is $\frac{x^2}{9} -\frac{y^2}{36}=1$.

Given:

There is a hyperbola with center at origin, the transverse axis along x-axis, eccentricity $\sqrt{5}$ and the sum of whose semi-axis is $9$.​

To Find:

The equation of the hyperbola with center at origin, the transverse axis along x-axis, eccentricity $\sqrt{5}$ and the sum of whose semi-axis is $9$.

Explanation:

The equation of the hyperbola with center at origin, the transverse axis along x-axis is as follows.

$\frac{x^2}{a^2} -\frac{y^2}{b^2}=1$

Note the following important points.

• The length of the major axis of the hyperbola is $=2a$
• The length of the minor axis of the hyperbola is $=2b$
• The distance between the two foci of the hyperbola is $=2c$
• The eccentricity of the hyperbola is $=e$

The following relationships are note-worthy.

• $e=\frac{c}{a}$
• $c^2=a^2+b^2$

Step 1 of 5

Mathematically we can get the following equations from the given conditions.

$e=\sqrt{5}$

$a+b=9$

Squaring both sides of $a+b=9$, we get the following.

$(a+b)^2=9^2\\a^2+2ab+b^2=81$

Step 2 of 5

Combining $e=\frac{c}{a}$ and $e=\sqrt{5}$, we get the following.

$\sqrt{5} =\frac{c}{a} \\5=\frac{c^2}{a^2} \\c^2=5a^2$

Step 3 of 5Step 4 of 5

Combining $c^2=a^2+b^2$ and $c^2=5a^2$, we get the following.

$5a^2=a^2+b^2\\b^2=4a^2=(2a)^2\\b=2a$

Step 4 of 5

Substituting the value of $b=2a$ in $a^2+2ab+b^2=81$, we get the following.

$a^2+2ab+b^2=81\\a^2+2a.(2a)+(2a)^2=81\\a^2+4a^2+4a^2=81\\9a^2=81\\a^2=9\\a=3$

Putting the value of $a=3$ in $a+b=9$, we get the following.

$3+b=9\\b=9-3\\b=6$

Step 5 of 5

Putting the value of $a=3,b=6$, the equation of the hyperbola is as follows.

$\frac{x^2}{3^2} -\frac{y^2}{6^2}=1\\\frac{x^2}{9} -\frac{y^2}{36}=1$

Therefore, the required equation of the hyperbola with center at origin, the transverse axis along x-axis, and the eccentricity $\sqrt{5}$  where the sum of the semi-axes is $9$ is $\frac{x^2}{9} -\frac{y^2}{36}=1$.

Know more from the following links.

https://brainly.in/question/337339

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