Question

Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbola x^2/16 - y^2/9 = 1

Answer 1

$\sf{\underline{We\:have\:to\:find:}}$

(i) Co-ordinates of the foci

(ii) Co-ordinates of the vertices

(iii) Eccentricity

(iv) Length of latus rectum


(i) $\sf{\underline{To\:find\:the\:co-ordinates of the vertices:}}$

$\sf{\underline{Equation\:of\:hyperbola:}}$

$\boxed{ \frac{ {x}^{2} }{16} - \frac{ {y}^{2} }{9} = 1}$


$\sf{\underline{In\:the\:form:}}$

$\boxed{ \frac{ {x}^{2} }{ {a}^{2} } - \frac{ {y}^{2} }{ {b}^{2} } = 1}$


Since, foci and vertices of the hyperbola lies on x-axis,

$\sf{\underline{Therefore:}}$

${a}^{2} = 16$, $a = \sqrt{16}$, $\boxed{a = 4}$

${b}^{2} = 9$, $b = \sqrt{9}$, $\boxed{b = 3}$

Now, we know the values of a and b.


To find the value of c, we have to use this formula:

$\boxed{ {c}^{2} = {a}^{2} + {b}^{2}}$


Now, substituting the above values in this formula, we get,

$\implies$ ${c}^{2} = {a}^{2} + {b}^{2}$

$\implies$ ${c}^{2} = 16 + 9$

$\implies$ ${c}^{2} = 25$

$\implies$ $c = \sqrt{25}$

$\implies$ $\boxed{c = 5}$

$\sf{\underline{We\:know\:that:}}$

$a = 4$, $b = 3$, $c = 5$


(i) $\sf{\underline{To\:find\:the\:co-ordinates\:of\:the\:foci:}}$

To find the coordinates of foci, we have to use this formula:

$\boxed{({\pm{c, 0}})}$

$\sf{\underline{Therefore:}}$

Co-ordinates of foci are

$\implies$ $({\pm{c, 0}})$

$\implies$ $\boxed{({\pm{5, 0}})}$


(ii) $\sf{\underline{To\:find\:the\:co-ordinates\:of\:the\:vertices:}}$

To find the coordinates of vertices, we have to use this formula:

$\boxed{({\pm{a, 0}})}$

$\sf{\underline{Therefore:}}$

Co-ordinates of vertices are

$\implies$ $({\pm{a, 0}})$

$\implies$ $\boxed{({\pm{4, 0}})}$


(iii) $\sf{\underline{To\:find\:the\:eccentricity:}}$

To find the eccentricity, we have to use this formula:

$\boxed{\frac{c}{a}}$

$\sf{\underline{Therefore:}}$

The eccentricity is

$\implies$ $\frac{c}{a}$

$\implies$ $\boxed{ \frac{5}{4}}$


(iv) $\sf{\underline{To\:find\:the\:length\:of\:latus\:rectum:}}$

To find the length of latus rectum, we have to use this formula:

$\boxed{ \frac{2b ^{2} }{a}}$

$\sf{\underline{Therefore:}}$

Length of latus rectum is

$\implies$ $\frac{ {2b}^{2} }{a}$

$\implies$ $\frac{2 \times 9}{4}$

$\implies$ $\boxed{\frac{9}{2}}$


$\sf{\underline{Therefore:}}$

(i) Co-ordinates of the foci

$\implies$ $\boxed{({\pm{5, 0}})}$

(ii) Co-ordinates of the vertices

$\implies$ $\boxed{({\pm{4, 0}})}$

(iii) Eccentricity

$\implies$ $\boxed{ \frac{5}{4}}$

(iv) Length of latus rectum

$\implies$ $\boxed{ \frac{9}{2}}$