00:01
Now, in this question, we have to graph the given hyperbola and also we have to specify the vertices, foci, length of transverse, conjugated axis and so on.
00:10
So, the given equation of hyperbola we have 16x square minus 32x minus 9y square plus 90y minus 353 is equals to 0.
00:31
Now, this equation is not in the standard form.
00:33
So, first we will make it in the standard form and for that i am going to add 353 to both side of the equation like this.
00:44
So we get 16x square minus 32x minus 9y square plus 90y this would become 0 is equals to 353.
01:00
Now, what i am doing here, i am writing brackets here this is 16x square minus 32x.
01:10
One more thing i can do, i can write 16x square as 4x whole square.
01:17
I can write in this way.
01:19
It's one and the same thing.
01:22
32x can be written as 2 times 4x times 4.
01:30
Now, i am adding 4 square here.
01:35
So, to the right side also we have to add 4 square.
01:38
So 353 plus 4 square.
01:41
So that there is no difference in equation.
01:47
Next.
01:47
So what i am doing here, i am taking a negative sign common here and 9 square can be written as 3y whole square like this plus 90y can be written as 2 times 3y times 15.
02:07
Now, i am adding 15 square here.
02:12
So plus 15 square.
02:14
Now here we have taken minus sign common.
02:16
So here we have to subtract 15 square.
02:18
So that equation is balanced.
02:20
Now, if you see this is what, this is the formula for 4x minus 4 whole square minus and this is the formula for 3y minus 15 whole square like this and this would be equals to 144.
02:40
Now, do one thing, divide with 144 both sides.
02:44
Here also 144, here also 144 and here also 144.
02:52
So what can i do here? now from the numerator, i can take 4 square common.
02:59
So we will get x minus 1 whole square divide by 144 and here i can take 3 square common.
03:07
So it would become y minus 5 whole square divide by 144 and this is equals to 1.
03:15
Now on further simplification, you will get x minus 1 whole square divide by 9 and 9 can be written as 3 square minus y minus 5 whole square divide by 16 and 16 can be written as 4 square is equals to 1.
03:39
Now if you recall the standard form of hyperbola, it is x minus h whole square divide by a square minus y minus k whole square divide by b square is equals to 1.
03:56
Now if you compare it with the equation that we have formed, we can say that here a is 3, b is 4, h is 1 and k is 5.
04:12
Now we can specify all those things that is asked in the equation which is like vertices, the eccentricity and all that stuff.
04:21
So first i am going to write the vertices.
04:23
So vertices for this kind of hyperbola is defined as h plus minus a comma k.
04:33
So i can write it here, vertices.
04:37
H we have 1 plus minus a.
04:43
So first i am taking positive sign.
04:45
So 1 plus 3 is 4 and k is 5.
04:52
Next i am taking negative sign here.
04:54
So 1 minus 3 is minus 2.
04:56
So minus 2 comma 5.
04:58
Next i can define the foci.
05:00
So foci we have is for this kind of hyperbola it is h plus minus c comma k.
05:10
But here we don't have the value of c.
05:11
So for that we know that c is square root a square plus b square.
05:19
A square is 9, b square is 16...