Question

Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$

Answer 1

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$\frac{x^2}{16}-\frac{y^2}{9}=1$ .......$\left(1\right)$

The above equation is of the form

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

So, axis of hyperbola is $x-$axis,

Comparing $\left(1\right)$ and $\left(2\right)$ we have

$a^2=16\Rightarrow a=4$ and $b^2=9\Rightarrow b=3$

Now, $c^2=a^2+b^2=16+9=25$

$\therefore c=\sqrt{25}=5$

Coordinate of foci$=(\pm c,0)=(\pm 5,0)$

Hence co-ordinates of foci are $(5,0)$ and $(-5,0)$

Co-ordinate of vertices$=(\pm a,0)=(\pm 4,0)$

So, Co-ordinate of vertices are $(4,0)$ and $(-4,0)$

Eccentricity$e=\frac{c}{a}=\frac{5}{4}$

latus rectum$=\frac{2b^2}{a}=\frac{2\times 9}{4}=\frac{9}{2}$

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