Hyperbola MCQ Questions & Answers in Geometry | Maths

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 $$\ell $$

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, $$O{M^2} + A{M^2} = O{A^2}$$
$$\eqalign{ & \therefore \,\frac{{{a^2}\left( {{b^2} + {\ell ^2}} \right)}}{{{b^2}}} + {\ell ^2} = 4{\ell ^2} \cr & {\text{we get }}{\ell ^2} = \frac{{{a^2}{b^2}}}{{3{b^2} - {a^2}}} \cr & {\text{Since, }}{\ell ^2} > 0 \cr & \therefore \frac{{{a^2}{b^2}}}{{3{b^2} - {a^2}}} > 0 \cr & \Rightarrow 3{b^2} - {a^2} > 0 \cr & \Rightarrow 3{a^2}\left( {{e^2} - 1} \right) - {a^2} > 0 \cr & \Rightarrow e > \frac{2}{{\sqrt 3 }} \cr} $$

Option A, $${x^2} + 2{y^2} \leqslant 1$$
It can be written as $$\frac{{{x^2}}}{1} + \frac{{{y^2}}}{{\frac{1}{2}}} \leqslant 1$$
which represents inside region of an ellipse
We know for an ellipse, the mid-point of any two points in the region, is also in the region.
Let's take two points $$\left( {0,\,0} \right)$$  and $$\left( {\frac{1}{2},\,0} \right)$$  which lies in the region.
Now, mid-point is $$\left( {\frac{1}{4},\,0} \right)$$  which also lies in the region.
Now, option B, $${\text{max }}\left\{ {\left| x \right|,\left| y \right|} \right\} \leqslant 1$$
Now, option C, $${x^2} - {y^2} \leqslant 1$$
which represents inside region of hyperbola.
A hyperbola has two parts . If we take 2 points , one in one part and other in other part, the mid-point need not to be inside the hyperbola.
Let's take $$\left( { - \frac{1}{2},\,0} \right)$$  and $$\left( {\frac{1}{4},\,0} \right)$$  as two points.
Their mid-point is $$\left( { - \frac{1}{8},\,0} \right)$$  is not in the region of hyperbola.
Hence, option C does not satisfy property P.
Option D, $${y^2} - {x^2} \leqslant 0$$
which represents inside region of pair of straight lines.
Mid-point of any two point in the region will lie in the region only.
Hence, satisfies property P.

Equation of hyperbola is $$\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$$
foci is $$\left( { \pm 2,\,0} \right)\,\, \Rightarrow ae = 2\,\, \Rightarrow {a^2}{e^2} = 4$$
$$\eqalign{ & {\text{ Since }}{b^2} = {a^2}\left( {{e^2} - 1} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{b^2} = {a^2}{e^2} - {a^2} \cr & \therefore {a^2} + {b^2} = 4.....(1) \cr} $$
Hyperbola passes through $$\left( {\sqrt 2 ,\,\sqrt 3 } \right)$$
$$\eqalign{ & \therefore \frac{2}{{{a^2}}} - \frac{3}{{{b^2}}} = 1.....(2) \cr & \Rightarrow \frac{2}{{4 - {b^2}}}\frac{{ - 3}}{{{b^2}}} = 1 \cr & \Rightarrow {b^4} + {b^2} - 12 = 0 \cr & \Rightarrow \left( {{b^2} - 3} \right)\left( {{b^2} + 4} \right) = 0 \cr & \Rightarrow {b^2} = 3 \cr & \,\,\,\,\,\,\,\,{b^2} = - 4\left( {{\text{Not possible}}} \right) \cr & {\text{For }}{b^2} = 3 \cr & \Rightarrow {a^2} = 1 \cr & \therefore \frac{{{x^2}}}{1} - \frac{{{y^2}}}{3} = 1 \cr} $$
Equation of tangent is $$\frac{{\sqrt 2 x}}{1} - \frac{{\sqrt 3 y}}{3} = 1$$
Clearly $$\left( {2\sqrt 2 ,\,3\sqrt 3 } \right)$$   satisfies it.

Let $$P\left( {x,\,y} \right)$$  be any point on the hyperbola and $$PM$$  is perpendicular from $$P$$ on the directrix,
Then by definition, $$SP = e\,PM$$
$$\eqalign{ & \Rightarrow {\left( {SP} \right)^2} = {e^2}{\left( {PM} \right)^2} \cr & \Rightarrow {\left( {x - 1} \right)^2} + {\left( {y - 2} \right)^2} = 3{\left\{ {\frac{{2x + y - 1}}{{\sqrt {4 + 1} }}} \right\}^2}\,\,\,\,\,\left( {\because \,\,e = \sqrt 3 } \right) \cr & \Rightarrow 5\left( {{x^2} + {y^2} - 2x - 4y + 5} \right) = 3\left( {4{x^2} + {y^2} + 1 + 4xy - 2y - 4x} \right) \cr & \Rightarrow 7{x^2} - 2{y^2} + 12xy - 2x + 14y - 22 = 0 \cr} $$
Which is the required hyperbola.

For hyperbola $$\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1,$$    we have
$$\frac{{2x}}{{{a^2}}} - \frac{{2y}}{{{b^2}}}\frac{{dy}}{{dx}} = 0\,\,\, \Rightarrow \frac{{dy}}{{dx}} = \frac{{{b^2}x}}{{{a^2}y}}$$
$$\therefore $$ Slope of normal at $$P\left( {6,\,3} \right)$$
$$ = - \frac{1}{{{{\left( {\frac{{dy}}{{dx}}} \right)}_{\left( {6,\,3} \right)}}}} = - \frac{{3{a^2}}}{{6{b^2}}}$$
$$\therefore $$ Equation of normal is $$\frac{{y - 3}}{{x - 6}} = - \frac{{3{a^2}}}{{6{b^2}}}$$
As it intersects $$x$$-axis at $$\left( {9,\,0} \right)$$
$$\therefore \frac{{0 - 3}}{{9 - 6}} = \frac{{ - 3{a^2}}}{{6{b^2}}}\,\,\, \Rightarrow {a^2} = 2{b^2}.....(1)$$
Also for hyperbola, $${b^2} = {a^2}\left( {{e^2} - 1} \right)$$
Using $${a^2} = 2{b^2};$$   we get
$$\eqalign{ & {b^2} = 2{b^2}\left( {{e^2} - 1} \right) \cr & \frac{1}{2} = {e^2} - 1 \cr & {\text{or,}}\,\,{e^2} = \frac{3}{2} \cr & {\text{or,}}\,\,e = \sqrt {\frac{3}{2}} \cr} $$

The equation of normal to the hyperbola $$\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$$    in terms of slope $$'m'$$ is $$y = mx \pm \frac{{m\left( {{a^2} + {b^2}} \right)}}{{\sqrt {{a^2} - {b^2}{m^2}} }}\,;$$
$${\text{Given line }}y = mx + \frac{{25\sqrt 3 }}{3}$$
and conic $$\frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{9} = 1$$    which is hyperbola with $${a^2} = 16,\,{b^2} = 9$$
By comparing given line with equation of normal we get
$$\eqalign{ & \pm \frac{{m\left( {{a^2} + {b^2}} \right)}}{{\sqrt {{a^2} - {b^2}{m^2}} }} = + \frac{{25\sqrt 3 }}{3} \cr & \Rightarrow \frac{{m\left( {16 + 9} \right)}}{{\sqrt {16 - 9{m^2}} }} = - \frac{{25\sqrt 3 }}{3} \cr & \Rightarrow \frac{{25m}}{{\sqrt {16 - 9{m^2}} }} = - \frac{{25\sqrt 3 }}{3} \cr & \Rightarrow 9{m^2} = 3\left( {16 - 9{m^2}} \right) \cr & \Rightarrow {m^2} = \frac{{16}}{{12}} \cr & \Rightarrow {m^2} = \frac{4}{3} \cr & \Rightarrow m = \pm \frac{2}{{\sqrt 3 }} \cr} $$

Equation of tangent to hyperbola $${x^2} - 2{y^2} = 4$$   at any point $$\left( {{x_1},\,{y_1}} \right)$$  is $$x{x_1} - 2y{y_1} = 4$$
Comparing with $$2x + \sqrt 6 y = 2$$    or $$4x + 2\sqrt 6 y = 4$$
$$ \Rightarrow {x_1} = 4$$   and $$ - 2{y_1} = 2\sqrt 6 \Rightarrow \left( {4,\, - \sqrt 6 } \right)$$      is the required point.

$$\eqalign{ & {\text{The tangent at}}\left( {2\sec \,\phi ,\,2\tan \,\phi } \right){\text{is }}x\sec \,\phi - y\tan \,\phi = 2 \cr & \therefore {a_1} = 2\cos \,\phi ,\,\,{b_1} = - 2\cot \,\phi \cr & {\text{The normal at}}\left( {2\sec \,\phi ,\,2\tan \,\phi } \right){\text{is }}x\cos \,\phi + y\cot \,\phi = 8 \cr & \therefore \,\,{a_2} = 8\sec \,\phi ,\,\,{b_2} = 8\tan \,\phi \cr & \therefore \,\,{a_1}{a_2} + {b_1}{b_2} = 16\cos \,\phi \,\sec \,\phi + \left( { - 2\cot \,\phi } \right)\left( {8\tan \,\phi } \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 16 - 16 = 0 \cr} $$

The length of transverse axis $$ = 2\,\sin \,\theta = 2a\,\, \Rightarrow a = \sin \,\theta $$
Also for ellipse $$3{x^2} + 4{y^2} = 12$$
$$\eqalign{ & {\text{or}}\,\,\,\,\,\frac{{{x^2}}}{4} + \frac{{{y^2}}}{3} = 1,\,\,{a^2} = 4,\,\,{b^2} = 3 \cr & e = \sqrt {1 - \frac{{{b^2}}}{{{a^2}}}} = \sqrt {1 - \frac{3}{4}} = \frac{1}{2} \cr} $$
$$\therefore $$ Focus of ellipse $$ = \left( {2 \times \frac{1}{2},\,0} \right) \Rightarrow \left( {1,\,0} \right)$$
As hyperbola is confocal with ellipse, focus of hyperbola $$ = \left( {1,\,0} \right) \Rightarrow ae = 1 \Rightarrow \sin \,\theta \times e = 1 \Rightarrow e = {\text{cosec}}\,\theta $$
$$\therefore {b^2} = {a^2}\left( {{e^2} - 1} \right) = {\sin ^2}\theta \left( {{\text{cose}}{{\text{c}}^2}\theta - 1} \right) = {\cos ^2}\theta $$
$$\therefore $$ Equation of hyperbola is
$$\eqalign{ & \frac{{{x^2}}}{{{{\sin }^2}\theta }} - \frac{{{y^2}}}{{{{\cos }^2}\theta }} = 1 \cr & {\text{or, }}{x^2}{\text{cose}}{{\text{c}}^2}\theta - {y^2}{\sec ^2}\theta = 1 \cr} $$

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