Question

Chap 5

Answer 1

Show that equation 16x^2-9y^2-96x-18y-9=0
represents a hyperbola.Determine the coordinates of the centre, the vertices, the equations at the asymptotes and the directrices.Sketch the hyperbola.

Answer 2

@kirbykirby

Answer 3

@rational

Answer 4

@ganeshie8

Answer 5

@kirbykirby

Answer 6

The hyperbola has the form:
\[ \frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1 \]
So you somehow have to manipulate your equation to get it into this form.

Answer 7

Since you have something of this form \((x-h)^2=x^2-2xh+h^2\) , (and the same for \(y-k)^2\), you notice that you have terms in \(x^2, x, y^2, y\) but you're missing the constant terms to make a perfect square. So usually you will have to use the complete the square method.

\[16x^2-9y^2-96x-18y-9=0\\
16x^2-96x-9y^2-18y=9 \]
Try and factor the leading coefficients of \(x^2\) and \(y^2\), and then complete the square on the resulting polynomial in ( ... ):
\[ 16(x^2-6x)-9(y^2+2y)=9\\
16(x^2-6x+9-9)-9(y^2+2y+1-1)=9\\
16\left[ (x-3)^2-9\right]-9 \left[ (y+1)^2-1\right]=9\\
16(x-3)^2-144-9(y+1)^2+9=9\\
16(x-3)^2 -9(y+1)^2 =144\\
\\~ \\
\frac{16(x-3)^2}{144}-\frac{9(y+1)^2}{144}=1\\
~ \\\dfrac{(x-3)^2}{\dfrac{144}{16}}-\dfrac{(y+1)^2}{\dfrac{144}{9}}=1\]