Question

The hyperbola $$\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$$    passes through the point $$\left( {3\sqrt 5 ,\,1} \right)$$  and the length of its latus rectum is $$\frac{4}{3}$$ units. The length of the conjugate axis is :

A. $$2$$ units
B. $$3$$ units
C. $$4$$ units  
D. $$5$$ units
Answer :   $$4$$ units
Solution :
$$\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$$
Hyperbola passes through $$\left( {3\sqrt 5 ,\,1} \right)$$
$$\eqalign{ & \therefore \,\frac{{{{\left( {3\sqrt 5 } \right)}^2}}}{{{a^2}}} - \frac{1}{{{b^2}}} = 1 \cr & \Rightarrow \frac{{45}}{{{a^2}}} - \frac{1}{{{b^2}}} = 1......\left( {\text{i}} \right) \cr} $$
Now length of latus rectum $$ = \frac{{2{b^2}}}{a}$$
$$\eqalign{ & \Rightarrow \frac{4}{3} = \frac{{2{b^2}}}{a} \cr & \Rightarrow \frac{2}{3} = \frac{{{b^2}}}{a} \cr & \Rightarrow a = \frac{{3{b^2}}}{2}......\left( {{\text{ii}}} \right) \cr} $$
Putting the value of $$'a'$$ from equation $$\left( {{\text{i}}} \right)$$ in equation $$\left( {{\text{ii}}} \right),$$
$$\eqalign{ & \Rightarrow \frac{{45 \times 4}}{{9{b^4}}} - \frac{1}{{{b^2}}} = 1 \cr & \Rightarrow \frac{{20}}{{{b^4}}} - \frac{1}{{{b^2}}} = 1 \cr & \Rightarrow 20 - {b^2} = {b^4} \cr & \Rightarrow {b^4} + {b^2} - 20 = 0 \cr & \Rightarrow {b^4} + 5{b^2} - 4{b^2} - 20 = 0 \cr & \Rightarrow {b^2}\left( {{b^2} + 5} \right) - 4\left( {{b^2} + 5} \right) = 0 \cr & \Rightarrow \left( {{b^2} - 4} \right)\left( {{b^2} + 5} \right) = 0 \cr & \Rightarrow {b^2} = 4,\,\,{b^2} = - 5 \cr & \therefore \,{b^2} = 4 \Rightarrow b = 2 \cr} $$
Now length of conjugate axis $$ = 2b = 2\left( 2 \right) = 4$$

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

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