let 0 theta pi 2 if the eccentricity of the hyperbola x

Let $$0 < \theta < \frac{\pi }{2}.$$ If the eccentricity of the hyperbola $$\frac{{{x^2}}}{{{{\cos }^2}\theta }} - \frac{{{y^2}}}{{{{\sin }^2}\theta }} = 1$$ is greater than $$2$$, then the length of its latus rectum lies in the interval :

A. $$\left( {3,\,\infty } \right)$$

B. $$\left( {\frac{3}{2},\,2} \right]$$

C. $$\left( {2,\,3} \right]$$

D. $$\left( {1,\,\frac{3}{2}} \right]$$

Answer : $$\left( {3,\,\infty } \right)$$

Solution :
$$\eqalign{ & \because {a^2} = {\cos ^2}\theta ,\,\,{b^2} = {\sin ^2}\theta \cr & {\text{and }}e > 2 \Rightarrow {e^2} > 4\,\, \Rightarrow 1 + \frac{{{b^2}}}{{{a^2}}}\,\, \Rightarrow 4 \cr & \Rightarrow 1 + {\tan ^2}\theta > 4 \cr & \Rightarrow {\sec ^2}\theta > 4 \Rightarrow \theta \Rightarrow \in \left( {\frac{\pi }{3},\,\frac{\pi }{2}} \right) \cr & {\text{Latus rectum,}} \cr & LR = \frac{{2{b^2}}}{a} = \frac{{2\,{{\sin }^2}\theta }}{{\cos \,\theta }} = 2\left( {\sec \,\theta - \cos \,\theta } \right) \cr & \Rightarrow \frac{{d\left( {LR} \right)}}{{d\theta }} = 2\left( {\sec \,\theta \,\tan \,\theta + \sin \,\theta } \right) > \forall \,\theta \, \in \left( {\frac{\pi }{3},\,\frac{\pi }{2}} \right) \cr & \therefore \min \left( {LR} \right) = 2\left( {\sec \frac{\pi }{3} - \cos \frac{\pi }{3}} \right) = 2 \left( {2 - \frac{1}{2}} \right) = 3 \cr} $$
max $$\left( {LR} \right)$$ tends to infinity as $$\theta \to \frac{\pi }{2}$$
Hence, length of latus rectum lies in the interval $$\left( {3,\,\infty } \right).$$

Releted MCQ Question on
Geometry >> Hyperbola

Releted Question 1

Each of the four inequalities given below defines a region in the $$xy$$ plane. One of these four regions does not have the following property. For any two points $$\left( {{x_1},\,{y_1}} \right)$$ and $$\left( {{x_2},\,{y_2}} \right)$$ in the the region, the point $$\left( {\frac{{{x_1} + {x_2}}}{2},\,\frac{{{y_1} + {y_2}}}{2}} \right)$$ is also in the region. The inequality defining this region is :

A. $${x^2} + 2{y^2} \leqslant 1$$

B. $${\text{max }}\left\{ {\left| x \right|,\left| y \right|} \right\} \leqslant 1$$

C. $${x^2} - {y^2} \leqslant 1$$

D. $${y^2} - {x^2} \leqslant 0$$

View Answer

Releted Question 2

Let $$P\left( {a\,\sec \,\theta ,\,b\,\tan \,\theta } \right)$$ and $$Q\left( {a\,\sec \,\phi ,\,b\,\tan \,\phi } \right),$$ where $$\theta + \phi = \frac{\pi }{2},$$ be two points on the hyperbola $$\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1.$$ If $$\left( {h,\,k} \right)$$ is the point of intersection of the normal at $$P$$ and $$Q,$$ then $$k$$ is equal to :

A. $$\frac{{{a^2} + {b^2}}}{a}$$

B. $$ - \left( {\frac{{{a^2} + {b^2}}}{a}} \right)$$

C. $$\frac{{{a^2} + {b^2}}}{b}$$

D. $$ - \left( {\frac{{{a^2} + {b^2}}}{b}} \right)$$

View Answer

Releted Question 3

If $$x=9$$ is the chord of contact of the hyperbola $${x^2} - {y^2} = 9,$$ then the equation of the corresponding pair of tangents is :

A. $$9{x^2} - 8{y^2} + 18x - 9 = 0$$

B. $$9{x^2} - 8{y^2} - 18x + 9 = 0$$

C. $$9{x^2} - 8{y^2} - 18x - 9 = 0$$

D. $$9{x^2} - 8{y^2} + 18x + 9 = 0$$

View Answer

Releted Question 4

For hyperbola $$\frac{{{x^2}}}{{{{\cos }^2}\alpha }} - \frac{{{y^2}}}{{{{\sin }^2}\alpha }} = 1,$$ which of the following remains constant with change in $$'\alpha \,'$$

A. abscissae of vertices

B. abscissae of foci

C. eccentricity

D. directrix

View Answer

Let $$0 < \theta < \frac{\pi }{2}.$$ If the eccentricity of the hyperbola $$\frac{{{x^2}}}{{{{\cos }^2}\theta }} - \frac{{{y^2}}}{{{{\sin }^2}\theta }} = 1$$ is greater than $$2$$, then the length of its latus rectum lies in the interval :

Releted MCQ Question on
Geometry >> Hyperbola

If $$x=9$$ is the chord of contact of the hyperbola $${x^2} - {y^2} = 9,$$ then the equation of the corresponding pair of tangents is :

For hyperbola $$\frac{{{x^2}}}{{{{\cos }^2}\alpha }} - \frac{{{y^2}}}{{{{\sin }^2}\alpha }} = 1,$$ which of the following remains constant with change in $$'\alpha \,'$$

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Hyperbola

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Let $$0 < \theta < \frac{\pi }{2}.$$ If the eccentricity of the hyperbola $$\frac{{{x^2}}}{{{{\cos }^2}\theta }} - \frac{{{y^2}}}{{{{\sin }^2}\theta }} = 1$$ is greater than $$2$$, then the length of its latus rectum lies in the interval :

Releted MCQ Question on Geometry >> Hyperbola

If $$x=9$$ is the chord of contact of the hyperbola $${x^2} - {y^2} = 9,$$ then the equation of the corresponding pair of tangents is :

For hyperbola $$\frac{{{x^2}}}{{{{\cos }^2}\alpha }} - \frac{{{y^2}}}{{{{\sin }^2}\alpha }} = 1,$$ which of the following remains constant with change in $$'\alpha \,'$$

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