MATHS Fill In The Blanks[1 Marks ]

Solution:

Let equation of hyperbola is $-\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1$

Vertices are $(0,\pm\text{b})\therefore\text{b}=6$ and $\text{e}=\frac{5}{3}$

We know that $\text{e}=\sqrt{1+\frac{\text{a}^2}{\text{b}^2}}$

$\Rightarrow\frac{5}{3}=\sqrt{1+\frac{\text{a}^2}{36}}$

$\Rightarrow\frac{25}{9}=1+\frac{\text{a}^2}{36}$

$\Rightarrow\frac{\text{a}^2}{36}=\frac{25}{9}-1=\frac{16}{9}$

$\Rightarrow\text{a}^2=\frac{16}{9}\times36$

$\Rightarrow\text{a}^2=64$

So, the equation of the hyperbola is,

$\frac{-\text{x}^2}{64}+\frac{\text{y}^2}{36}=1$

$\Rightarrow\frac{\text{y}^2}{36}-\frac{\text{x}^2}{64}=1$

and $\text{foci}=(0,\pm\text{be})=\Big(0,\pm6\times\frac{5}{3}\Big)=(0,\pm10)$

Hence, the value of the filler is $\frac{\text{y}^2}{36}-\frac{\text{x}^2}{64}=1$ and $(0,\pm10)$ 

Solution:

Solution: