Hyperbola MCQ Questions & Answers in Geometry | Maths

The given equation of curve is
$$\eqalign{ & 4{x^2} - 9{y^2} = 1 \cr & \Rightarrow \frac{{{x^2}}}{{\frac{1}{4}}} - \frac{{{y^2}}}{{\frac{1}{9}}} = 1 \cr} $$
This is an equation of a hyperbola which does not intersect with conjugate axes.
Hence, no point of intersection exists.

For rectangular hyperbola, $$e = \sqrt 2 .$$  So, $${\left( {x - 1} \right)^2} + {\left( {y - 2} \right)^2} = {\left( {\sqrt 2 .\frac{{x + y - 1}}{{\sqrt 2 }}} \right)^2}.$$

$$\eqalign{ & x = 5\tan \,\phi ,\,\,y = \,4\sec \,\phi \, \Rightarrow \frac{{{y^2}}}{{16}} - \frac{{{x^2}}}{{25}} = 1\,\, \Rightarrow {a^2} = 16,\,{b^2} = 25 \cr & {\text{Also, }}{b^2} = {a^2}\left( {{e^2} - 1} \right) \cr} $$

The given equation of hyperbola is
$$\eqalign{ & \frac{{{x^2}}}{{{{\cos }^2}\alpha }} - \frac{{{y^2}}}{{{{\sin }^2}\alpha }} = 1 \cr & \Rightarrow a = \cos \,\alpha ,\,\,\,b = \sin \,\alpha \cr & \Rightarrow e = \sqrt {1 + \frac{{{b^2}}}{{{a^2}}}} = \sqrt {1 + {{\tan }^2}\alpha } = \sec \,\alpha \cr & \Rightarrow ae = 1 \cr} $$
$$\therefore $$ foci $$\left( { \pm 1,\,0} \right)$$
$$\therefore $$ foci remain constant with respect to $$\alpha .$$

The standard equation of hyperbola is $$\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$$
Latus rectum $$ = \frac{{2{b^2}}}{a},$$
Conjugate axis $$ = 2b,$$
Distance between the foci $$= 2 ae$$
According to the question,
$$\eqalign{ & \frac{{2{b^2}}}{a} = 8......\left( {\text{i}} \right) \cr & 2b = \frac{1}{2}\left( {2ae} \right) \Rightarrow b = \frac{{ae}}{2}......\left( {{\text{ii}}} \right) \cr} $$
From equation $$\left( {\text{i}} \right)\& \left( {{\text{ii}}} \right),$$
$$\eqalign{ & \frac{2}{a}{\left( {\frac{{ae}}{2}} \right)^2} = 8 \cr & \Rightarrow 2.\frac{{{a^2}{e^2}}}{{4a}} = 8 \cr & \Rightarrow a{e^2} = 16......\left( {{\text{iii}}} \right) \cr} $$
From equation $$\left( {\text{i}} \right),\,\,{b^2} = 4a$$
Using $${b^2} = {a^2}\left( {{e^2} - 1} \right)$$    we get
$$\eqalign{ & \left( {4a} \right) = {a^2}\left( {{e^2} - 1} \right) \cr & \Rightarrow 4 = \frac{{16}}{{{e^2}}}\left( {{e^2} - 1} \right) \cr & \Rightarrow 16 - \frac{{16}}{{{e^2}}} = 4 \cr & \Rightarrow \frac{{16}}{{{e^2}}} = 12 \cr & \therefore \,e = \frac{2}{{\sqrt 3 }}. \cr} $$

Equation of the normal to the hyperbola $$\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$$   at the point $$\left( {a\,\sec \,\alpha ,\,b\,\tan \,\alpha } \right)$$    is given by $$ax\,\cos \,\alpha + by\,\cot \,\alpha = {a^2} + {b^2}$$
Normals at $$\theta ,\,\phi $$  are
\[\left\{ \begin{array}{l} ax\,\cos \,\theta + by\,\cot \,\theta = {a^2} + {b^2}\\ ax\,\cos \,\phi + by\,\cot \,\phi = {a^2} + {b^2} \end{array} \right.\]
where $$\phi = \frac{\pi }{2} - \theta $$   and these passes through $$\left( {h,\,k} \right)$$
$$\eqalign{ & \therefore \,ah\,\cos \,\theta + bk\,\cot \,\theta = {a^2} + {b^2} \cr & \Rightarrow ah\,\sin \,\theta + bk\,\tan \,\theta = {a^2} + {b^2} \cr} $$
Eliminating $$h,\,\,bk\left( {\cot \,\theta \,\sin \,\theta - \tan \,\theta \,\cos \,\theta } \right)$$
$$ = \left( {{a^2} + {b^2}} \right)\left( {\sin \,\theta - \cos \,\theta } \right)\,{\text{or }}k = - \frac{{\left( {{a^2} + {b^2}} \right)}}{b}$$

Let the points lie on the circle $${x^2} + {y^2} + 2gx + 2fy + 2fy + k = 0,$$        then
$$\eqalign{ & {c^2}t_{\text{i}}^2 + \frac{{{c^2}}}{{t_{\text{i}}^2}} + 2gc{t_{\text{i}}} + 2f\frac{c}{{{t_{\text{i}}}}} + k = 0 \cr & \Rightarrow {c^2}t_{\text{i}}^4 + 2gct_{\text{i}}^3 + kt_{\text{i}}^3 + 2fc{t_{\text{i}}} + {c^2} = 0 \cr} $$
Its roots are $${t_1},\,{t_2},\,{t_3},\,{t_4}$$    so $${t_1}{t_2}{t_3}{t_4} = \frac{{{c^2}}}{{{c^2}}} = 1$$
Also, $${t_1} + {t_2} + {t_3} + {t_4} = - \frac{{2gc}}{{{c^2}}} = - \frac{{2g}}{c}$$

$$\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$$
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 $$ = 2b = 2\left( 2 \right) = 4$$

$$\eqalign{ & \frac{{2{b^2}}}{a} = 8{\text{ and }}2b = \frac{1}{2}\left( {2ae} \right) \cr & \Rightarrow 4{b^2} = {a^2}{e^2} \cr & \Rightarrow 4{a^2}\left( {{e^2} - 1} \right) = {a^2}{e^2} \cr & \Rightarrow 3{e^2} = 4 \cr & \Rightarrow e = \frac{2}{{\sqrt 3 }} \cr} $$

Chord $$x=9$$  meets $${x^2} - {y^2} = 9$$   at $$\left( {9,\,6\sqrt 2 } \right)$$  and $$\left( {9,\, - 6\sqrt 2 } \right)$$  at which tangents are
$$\eqalign{ & 9x - 6\sqrt 2 y = 9{\text{ and }}9x + 6\sqrt 2 y = 9 \cr & {\text{or }}3x - 2\sqrt 2 y - 3 = 0{\text{ and }}3x + 2\sqrt 2 y - 3 = 0 \cr} $$
$$\therefore $$ Combined equation of tangents is
$$\eqalign{ & \left( {3x - 2\sqrt 2 y - 3} \right)\left( {3x + 2\sqrt 2 y - 3} \right) = 0 \cr & {\text{or }}\,9{x^2} - 8{y^2} - 18x + 9 = 0 \cr} $$