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If the vertex of a hyperbola bisects the distance between its centere and the corresponding focus, then ratio of square of its conjugate axis of the square of its transverse axis is
- a)2
- b)4
- c)6
- d)3
Correct answer is option 'D'. Can you explain this answer?
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If the vertex of a hyperbola bisects the distance between its centere ...
a = ae, i.e.,
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If the vertex of a hyperbola bisects the distance between its centere ...
Given: The vertex of a hyperbola bisects the distance between its center and the corresponding focus.
To find: The ratio of the square of its conjugate axis to the square of its transverse axis.
Solution:
Let the hyperbola be of the form $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$.
Let the center of the hyperbola be $(0,0)$.
Let the distance between the center and the focus be $c$.
Let the vertex of the hyperbola be $(0,\pm a)$.
Then, the foci of the hyperbola are at $(0,\pm c)$.
Given that the vertex bisects the distance between the center and the focus, we have:
$$\frac{c}{2}=a$$
Solving for $c$, we get:
$$c=2a$$
The distance between the vertices of the hyperbola is $2a$.
The distance between the endpoints of the conjugate axis is $2b$.
Therefore, the ratio of the square of the conjugate axis to the square of the transverse axis is:
$$\left(\frac{2b}{2a}\right)^2=\left(\frac{b}{a}\right)^2$$
Substituting $c=2a$ in the equation of the hyperbola, we get:
$$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \Rightarrow \frac{x^2}{a^2}-\frac{y^2}{a^2-c^2}=1$$
Substituting $c=2a$, we get:
$$\frac{x^2}{a^2}-\frac{y^2}{a^2-4a^2}=1 \Rightarrow \frac{x^2}{a^2}-\frac{y^2}{-3a^2}=1$$
Therefore, $a^2=3b^2$.
Substituting this in the ratio of the square of the conjugate axis to the square of the transverse axis, we get:
$$\left(\frac{b}{a}\right)^2=\left(\frac{1}{\sqrt{3}}\right)^2=\frac{1}{3}$$
Therefore, the answer is option D, 3.
To find: The ratio of the square of its conjugate axis to the square of its transverse axis.
Solution:
Let the hyperbola be of the form $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$.
Let the center of the hyperbola be $(0,0)$.
Let the distance between the center and the focus be $c$.
Let the vertex of the hyperbola be $(0,\pm a)$.
Then, the foci of the hyperbola are at $(0,\pm c)$.
Given that the vertex bisects the distance between the center and the focus, we have:
$$\frac{c}{2}=a$$
Solving for $c$, we get:
$$c=2a$$
The distance between the vertices of the hyperbola is $2a$.
The distance between the endpoints of the conjugate axis is $2b$.
Therefore, the ratio of the square of the conjugate axis to the square of the transverse axis is:
$$\left(\frac{2b}{2a}\right)^2=\left(\frac{b}{a}\right)^2$$
Substituting $c=2a$ in the equation of the hyperbola, we get:
$$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \Rightarrow \frac{x^2}{a^2}-\frac{y^2}{a^2-c^2}=1$$
Substituting $c=2a$, we get:
$$\frac{x^2}{a^2}-\frac{y^2}{a^2-4a^2}=1 \Rightarrow \frac{x^2}{a^2}-\frac{y^2}{-3a^2}=1$$
Therefore, $a^2=3b^2$.
Substituting this in the ratio of the square of the conjugate axis to the square of the transverse axis, we get:
$$\left(\frac{b}{a}\right)^2=\left(\frac{1}{\sqrt{3}}\right)^2=\frac{1}{3}$$
Therefore, the answer is option D, 3.
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If the vertex of a hyperbola bisects the distance between its centere and the corresponding focus, then ratio of square of its conjugate axis of the square of its transverse axis isa)2b)4c)6d)3Correct answer is option 'D'. Can you explain this answer?
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