Question

If $$AB$$  is a double ordinate of the hyperbola $$\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$$    such that $$\Delta OAB$$   is an equilateral triangle $$O$$ being the origin, then the eccentricity of the hyperbola satisfies :

A. $$e > \sqrt 3 $$
B. $$1 < e < \frac{2}{{\sqrt 3 }}$$
C. $$e = \frac{2}{{\sqrt 3 }}$$
D. $$e > \frac{2}{{\sqrt 3 }}$$  
Answer :   $$e > \frac{2}{{\sqrt 3 }}$$
Solution :
Let the length of the double ordinate be $$2\ell $$
$$\therefore \,AB = 2\ell {\text{ and }}AM = BM = \ell $$
Clearly ordinate of point $$A$$ is $$\ell $$
Hyperbola mcq solution image
The abscissa of the point $$A$$ is given by
$$\eqalign{ & \frac{{{x^2}}}{{{a^2}}} - \frac{{{\ell ^2}}}{{{b^2}}} = 1 \Rightarrow x = \frac{{a\sqrt {{b^2} + {\ell ^2}} }}{b} \cr & \therefore \,A{\text{ is }}\left( {\frac{{a\sqrt {{b^2} + {\ell ^2}} }}{b},\,\ell } \right) \cr} $$
Since $$\Delta OAB$$   is equilateral triangle, therefore $$OA = AB = OB = 2\ell $$
Also, $$O{M^2} + A{M^2} = O{A^2}$$
$$\eqalign{ & \therefore \,\frac{{{a^2}\left( {{b^2} + {\ell ^2}} \right)}}{{{b^2}}} + {\ell ^2} = 4{\ell ^2} \cr & {\text{we get }}{\ell ^2} = \frac{{{a^2}{b^2}}}{{3{b^2} - {a^2}}} \cr & {\text{Since, }}{\ell ^2} > 0 \cr & \therefore \frac{{{a^2}{b^2}}}{{3{b^2} - {a^2}}} > 0 \cr & \Rightarrow 3{b^2} - {a^2} > 0 \cr & \Rightarrow 3{a^2}\left( {{e^2} - 1} \right) - {a^2} > 0 \cr & \Rightarrow e > \frac{2}{{\sqrt 3 }} \cr} $$

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. $${y^2} - {x^2} \leqslant 0$$
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. $$ - \left( {\frac{{{a^2} + {b^2}}}{b}} \right)$$
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. $$9{x^2} - 8{y^2} + 18x + 9 = 0$$
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

Practice More Releted MCQ Question on
Hyperbola


Practice More MCQ Question on Maths Section