Question

Find the equation of the hyperbola whose

(i) focus is at (5,2),vertex at (4,2) and centre at(3,2)

(ii) focus is at (4,2), centre at (6,2) and e=2.

Answer 1

$\text{(i) Let}(x_2,y_2)\text{be the coordinates of the second vertex}$

$\text{We know that the centre of the hyperbola is the mid-point of the line-joining the two vertices.}$

$\therefore \frac{x_1+4}{2}=3and\frac{y_1+2}{2}=2$

$[\because Centre=(3,2)andvertex=(4,2\left)\right]$

$\Rightarrow x_1=2and{y}_2=2$

$\therefore \text{The coordinates of the second vertex is (2,2)}$

$\text{Let 2a and 2b the length of gransvers and conjugate axes and let e be eccentreicity.Then the equation of hyperbola is}$

$\frac{(x-3{)}^2}{a^2}-\frac{(y-2{)}^2}{b^2}=1...\left(i\right)$

$\text{Now, distance between the two vertices}$

$=2a$ $\Rightarrow \sqrt{(4-2{)}^2+(2-2{)}^2}=2a$

$[\because vertices=(4,2\left)and\right(2,2\left)\right]$

$\Rightarrow \sqrt{(2{)}^2}=2a$ $\Rightarrow 2a=2$

$\Rightarrow a=a...\left(iii\right)$

$\text{Now,the distance between the vertex and focus is=ae-a}$

$\Rightarrow \sqrt{(5-4{)}^2+(2-2{)}^2}=ae-a$

$[\because Focus=(5,2\left)andvertex\right(4,2\left)\right]$

$\Rightarrow \sqrt{1}=ae-a$

$\Rightarrow ae-a=1$

$\to 1\times e-1=1[\because e=1]$ $\Rightarrow e=1+1=2$

$Now,$

$b^2=a^2(a^2-1)$

$=a^2(2^2-1)$

$=1\times (4-1)$

$=1\times 3$

$=3$

$Putting{a}^2=1and{b}^2=3inequation\left(i\right),weget$

$\Rightarrow \frac{(x-3{)}^2}{1}-\frac{(y-2{)}^2}{3}=1$

$\Rightarrow \frac{3(x-3{)}^2-(y-2{)}^2}{3}=1$

$\Rightarrow 3(x-3{)}^2-y(y-2{)}^2=3$

$\text{This is the equation of the required hyperbola.}$

$\left(ii\right)\text{Let}(x_1,y_1)\text{be the coordinates of the second vertex focus of the requires hyperbola.}$

$\text{We know that,the ventre of the hyperbola is the mid-point of the line-joining the two foci.}$

$\therefore \frac{x_1+4}{2}=6and\frac{y_1+2}{2}=2$

$[\because Centre=(6,2)andfocus=(4,2\left)\right]$

$\Rightarrow x_1=8and{y}_2=2$

$\therefore \text{The coordinates of the second focus is (8,2)}$

$\text{Let 2a and 2b be the length of transverse and conjugate axes and let e be eccentricity.Then the equation of hyperbola is}$

$\frac{(x-6{)}^2}{a^2}-\frac{(y-2{)}^2}{b^2}=1...\left(i\right)$

$\text{Now,distance between the two vertices=2ae}$

$\Rightarrow \sqrt{(8-4{)}^2+(2-2{)}^2}=2ae$

$[\because foci=(4,2\left)and\right(8,2\left)\right]$

$\Rightarrow \sqrt{(2{)}^2}=2a$

$\Rightarrow 2a=2$

$\Rightarrow a=1...\left(ii\right)$

$\text{Now,the distance between the vertex and focus is~=~ae~-~a}$

$\Rightarrow \sqrt{(5-4{)}^2+(2-2{)}^2}=ae-a$

$[\because Focus=(5,2\left)andvertex\right(4,2\left)\right]$

$\Rightarrow \sqrt{(5-4{)}^2+(2-2{)}^2}=ae-a$

$\Rightarrow \sqrt{(4{)}^2}=2ae$

$\Rightarrow 2ae=4$

$\Rightarrow 2\times a\times =4[\because e=2]$

$\Rightarrow a=1$

$\Rightarrow a^2=1$

$Now,$ $b^2=a^2(e^2-1)$

$\Rightarrow b^2=1(2^2-1)$

$=1(4-1)$ $=3$

$Putting{a}^2=1and{b}^2=3inequation\left(i\right),$

$weget$

$\Rightarrow \frac{(x-6{)}^2}{1}-\frac{(y-2{)}^2}{3}=1$

$\Rightarrow \frac{3(x-6{)}^2-(y-2{)}^2}{3}=1$

$\Rightarrow 3(x-6{)}^2-(y-2{)}^2=3$

$\text{This is the equaion of the required hyperbola}$