Hyperbola MCQ Questions & Answers in Geometry | Maths

Here, $$\frac{{{x^2}}}{1} - \frac{{{y^2}}}{4} = 1.$$   Any tangent to the hyperbola is $$y = mx \pm \sqrt {{a^2}{m^2} - {b^2}} ,$$     i.e., $$y = mx \pm \sqrt {{m^2} - 4} $$     in which $$m$$ is any real number.

Given, equation of hyperbola is $$\frac{{{x^2}}}{{{{\cos }^2}\alpha }} - \frac{{{y^2}}}{{{{\sin }^2}\alpha }} = 1$$
We know that the equation of hyperbola is $$\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$$
Here, $${a^2} = {\cos ^2}\alpha $$   and $${b^2} = {\sin ^2}\alpha $$
We know that, $${b^2} = {a^2}\left( {{e^2} - 1} \right)$$
$$\eqalign{ & \Rightarrow {\sin ^2}\alpha = {\cos ^2}\alpha \left( {{e^2} - 1} \right) \cr & \Rightarrow {\sin ^2}\alpha + {\cos ^2}\alpha = {\cos ^2}\alpha .{e^2} \cr & \Rightarrow {e^2} = 1 + {\tan ^2}\alpha = {\sec ^2}\alpha \cr & \Rightarrow e = \sec \,\alpha \cr & \therefore ae = \cos \,\alpha .\frac{1}{{\cos \,\alpha }} = 1 \cr} $$
Co-ordinates of foci are ($$\left( { \pm ae,\,0} \right)$$  i.e., $$\left( { \pm 1,\,0} \right)$$
Hence, abscissae of foci remain constant when $$\alpha $$ varies.

Given, hyperbola is
$$\eqalign{ & {\left( {10x - 5} \right)^2} + {\left( {10y - 2} \right)^2} = 9{\left( {3x + 4y - 7} \right)^2} \cr & \Rightarrow {\left( {x - \frac{1}{2}} \right)^2} + {\left( {y - \frac{1}{5}} \right)^2} = \frac{9}{4}{\left( {\frac{{3x + 4y - 7}}{5}} \right)^2} \cr} $$
$$ \Rightarrow $$  Given curve is a hyperbola where focus is $$\left( {\frac{1}{2},\,\frac{1}{5}} \right)$$  and directrix is $$3x + 4y - 7 = 0.$$    Latus rectum is a line passing through the focus and parallel to the directrix.
$$ \Rightarrow $$  Equation of the latus rectum is $$y - \frac{1}{5} = - \frac{3}{4}\left( {x - \frac{1}{2}} \right).$$

$$\eqalign{ & {a^2} - {b^2} = 512 \cr & \Rightarrow \left( {a + b} \right)\left( {a - b} \right) = {2^9} \cr & \Rightarrow \left( {a + b,\,a - b} \right) = \left( {{2^8},\,2} \right),\,\left( {{2^7},\,{2^2}} \right),\,\left( {{2^6},\,{2^3}} \right),\,\left( {{2^5},\,{2^4}} \right) \cr} $$
Since, $$a > b,\,a + b > a - b$$     therefore the other combinations like $$\left( {{2^4},\,{2^5}} \right)$$  etc cannot be accepted. $$\left( {{2^9},\,1} \right)$$  also cannot be accepted since $$a$$ and $$b$$ are positive integers.

Equation of the tangent at the point $$'\theta '$$ is $$\frac{{x\,\sec \,\theta }}{a} - \frac{{y\,\tan \,\theta }}{b} = 1$$
$$\eqalign{ & \Rightarrow A{\text{ is}}\left( {a\,\cos \,\theta ,\,0} \right){\text{and }}B{\text{ is}}\left( {0,\, - b\,\cot \,\theta } \right) \cr & {\text{Let }}P{\text{ be}}\left( {h,\,k} \right) \Rightarrow h = a\,\cos \,\theta ,\,k = - b\,\cot \,\theta \cr & \Rightarrow \frac{k}{h} = - \frac{b}{{a\,\sin \,\theta }} \cr & \Rightarrow \sin \,\theta = - \frac{{bh}}{{ak}}{\text{ and cos}}\,\theta = \frac{h}{a} \cr} $$

Square and add,
$$\eqalign{ & \Rightarrow \frac{{{b^2}{h^2}}}{{{a^2}{k^2}}} + \frac{{{h^2}}}{{{a^2}}} = 1 \cr & \Rightarrow \frac{{{b^2}}}{{{k^2}}} + 1 = \frac{{{a^2}}}{{{h^2}}} \cr & \Rightarrow \frac{{{a^2}}}{{{h^2}}} - \frac{{{b^2}}}{{{k^2}}} = 1 \cr} $$
Hence, locus of $$P$$ is $$\frac{{{a^2}}}{{{x^2}}} - \frac{{{b^2}}}{{{y^2}}} = 1$$

Equation of the ellipse is $$3{x^2} + 4{y^2} = 12$$
$$ \Rightarrow \frac{{{x^2}}}{4} + \frac{{{y^2}}}{3} = 1......\left( 1 \right)$$
Eccentricity $${e_1} = \sqrt {1 - \frac{3}{4}} = \frac{1}{2}$$
So, the foci of ellipse are $$\left( {1,\,0} \right)$$  and $$\left( { - 1,\,0} \right)$$
Let the equation of the required hyperbola be
$$\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1......\left( 2 \right)$$
Given $$2a = 2\,\sin \,\theta \Rightarrow a = \sin \,\theta $$
Since the ellipse $$\left( 1 \right)$$ and the hyperbola $$\left( 2 \right)$$ are confocal, so the foci of hyperbola are $$\left( {1,\,0} \right)$$  and $$\left( { - 1,\,0} \right)$$  too.
If the eccentricity, of hyperbola be $${e_2}$$ then
$$\eqalign{ & a{e_2} = 1 \Rightarrow \sin \,\theta {e_2} = 1 \Rightarrow {e_2} = {\text{cosec}}\,\theta \cr & \therefore \,{b^2} = {a^2}\left( {e_2^2 - 1} \right) = {\sin ^2}\theta \left( {{\text{cose}}{{\text{c}}^2}\theta - 1} \right) = {\cos ^2}\theta \cr} $$
$$\therefore $$  Required equation of the hyperbola is
$$\frac{{{x^2}}}{{{{\sin }^2}\theta }} - \frac{{{y^2}}}{{{{\cos }^2}\theta }} = 1 \Rightarrow {x^2}{\text{cose}}{{\text{c}}^2}\theta - {y^2}{\sec ^2}\theta = 1$$

Any tangent to the hyperbola is $$y = mx + \sqrt {{a^2}{m^2} - {b^2}} .$$      This is the same as $$y = \alpha x + \beta .$$
$$\eqalign{ & \therefore \,\,m = \alpha ,\,\sqrt {{a^2}{m^2} - {b^2}} \, = \beta \Rightarrow {a^2}{\alpha ^2} - {\beta ^2} = {b^2} \cr & \therefore \,\,{\text{the locus is }}{a^2}{x^2} - {y^2} = {b^2} \cr} $$