Question

NO LINKS!! Please help me with this problem​

Answer 1

Hello and Good Morning/Afternoon:

Let's solve this problem step-by-step:

Let's find the format of the standard form of hyperbole:

 $\hookrightarrow \frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2} =1$

• (h,k): hyperbole center ⇒(3,-5)

Let's find the value of a, b:

• value of a: distance between vertex (7, -5) and center (3, -5)

             $a = \sqrt{(7-3)^2+(-5--5)^2} =\sqrt{16} =4$

• value of b: $\sqrt{c^2-a^2}$

           ⇒value of c: distance between focus (9, -5) and center (3, -5)

                  $c = \sqrt{(3-9)^2+(-5--5)^2}=\sqrt{36} =6$

            ⇒ therefore:

                  $b = \sqrt{c^2-a^2} =\sqrt{6^2-4^2}=\sqrt{20}$

Let's plug everything into our standard form of the equation:

   $\frac{(x-3)^2}{4^2} -\frac{(y--5)^2}{(\sqrt{20})^2 } =1\\\frac{(x-3)^2}{16} -\frac{(y+5)^2}{20 } =1$ <== Answer

Hope that helps!

#LearnwithBrainly

Answer 2

Information : The given hyperbola is a horizontal hylerbola with its centre (3 , -5) and one of its focus at (9 , -5) and vertex at (7 , -5) and as we can see that the focus and vertex have same y - coordinates, it must have its Transverse axis on line y = - 5.

Now,

it's vertex is given, I.e (7 , -5)

so, length of semi transverse axis will be equal to distance of vertex from centre, i.e

• a = 7 - 3 = 4 units

Now, it's focus can be represented as ;

$\qquad \sf \dashrightarrow \: (3 + ae, - 5 )$

so,

• ae + 3 = 9

and we know, a = 4

$\qquad \sf \dashrightarrow \: 4e + 3 = 9$

$\qquad \sf \dashrightarrow \: 4e = 6$

$\qquad \sf \dashrightarrow \: e = \cfrac{3}{2}$

Now, let's find the measure of semi - conjugate axis (b)

$\qquad \sf \dashrightarrow \: {b}^{2} = {a}^{2} ( {e}^{2} - 1)$

$\qquad \sf \dashrightarrow \: {b}^{2} = 16( \frac{9}{4} - 1)$

$\qquad \sf \dashrightarrow \: {b}^{2} = 16( \frac{9 - 4}{4} )$

$\qquad \sf \dashrightarrow \: {b}^{2} = 16( \frac{5}{4} )$

$\qquad \sf \dashrightarrow \: {b}^{2} = 20$

$\qquad \sf \dashrightarrow \: b = \sqrt{20}$

So, it's time to write the equation of hyperbola, as we already have the values of a and b ~

$\qquad \sf \dashrightarrow \: \cfrac{ {(x - h)}^{2} }{ {a}^{2} } - \cfrac{( {y - k)}^{2} }{ {b}^{2} } = 1$

[ plug in the values, and h = x - coordinate of centre, and k = y - coordinate of centre ]

$\qquad \sf \dashrightarrow \: \cfrac{ ({x-3)}^{2} }{ {16}^{} } - \dfrac{ {(y+5)}^{2} }{ { {20} }^{} } = 1$