Respuesta :
24)
Given:
[tex]-9x^2+72x+16y^2+16y+4=0[/tex]We need to find the hyperbola in standard form
[tex]\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1[/tex]The given equation can be written as follows.
[tex]-9(x^2+8x)+16(y^2+y)+4=0[/tex][tex]-9(x^2+2\times4x+4^2-4^2)+16(y^2+2\times\frac{1}{2}y+\frac{1}{2^2}-\frac{1}{2^2})+4=0[/tex][tex]-9(x^2+2\times4x+4^2)-(-9)4^2+16(y^2+2\times\frac{1}{2}y+\frac{1}{2^2})-(16)\frac{1}{2^2}+4=0[/tex][tex]-9(x^2+2\times4x+4^2)+144+16(y^2+2\times\frac{1}{2}y+\frac{1}{2^2})-4+4=0[/tex][tex]-9(x^2+2\times4x+4^2)+144+16(y^2+2\times\frac{1}{2}y+\frac{1}{2^2})=0[/tex][tex]\text{ Use (a+b)}^2=a^2+2ab+b^2\text{.}[/tex][tex]-9(x+4)^2+144+16(y+\frac{1}{2})^2=0[/tex]Subtracting 144 from both sides, we get
[tex]-9(x+4)^2+144+16(y+\frac{1}{2})^2-144=0-144[/tex][tex]-9(x+4)^2+16(y+\frac{1}{2})^2=-144[/tex]Dividing both sides by (-144), we get
[tex]\frac{-9\mleft(x+4\mright)^2}{-144}+\frac{16(y+\frac{1}{2})^2}{-144}=-\frac{144}{-144}[/tex][tex]\frac{\mleft(x+4\mright)^2}{16}-\frac{(y+\frac{1}{2})^2}{9}=1[/tex][tex]\frac{\mleft(x+4\mright)^2}{4^2}-\frac{(y+\frac{1}{2})^2}{3^2}=1[/tex]Hence we get the standard form of the hyperbola.
[tex]\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1[/tex]Comapring with standerd form, we get
[tex]h=-4,k=-\frac{1}{2},a=4\text{ and b=3.}[/tex]The coordinates of the vertices are
[tex](h+a,k)\text{ and }(h-a,k)\text{ }[/tex]Substitute h=-4, a=4 and k=-1/2.
[tex](-4+4,-\frac{1}{2})\text{ and }(-4-4,-\frac{1}{2})[/tex][tex](0,-\frac{1}{2})\text{ and }(-8,-\frac{1}{2})[/tex][tex](0,-0.5)\text{ and }(-8,-0.5)[/tex]Hence the vertices are (0, -0.5) and (-8, -0.5).
The distance between foci is 2c,
[tex]c^2=a^2+b^2[/tex]Substitute a=4 and b=3 to find the value of c.
[tex]c^2=4^2+3^2[/tex][tex]c^2=16+9=25^{}[/tex][tex]c=\pm\sqrt[]{25}[/tex][tex]c=\pm5[/tex]The measure of distance can not be negative.
[tex]c=5[/tex]The coordinates of the foci are
[tex](h+c,k)\text{ and }(h-c,k)[/tex]Substitute h=-4, k=-1/2 and c=5, we get
[tex](-4+5,-\frac{1}{2})\text{ and }(-4-5,-\frac{1}{2})[/tex][tex](1,-\frac{1}{2})\text{ and }(-9,-\frac{1}{2})[/tex][tex](1,-0.5)\text{ and }(-9,-0.5)[/tex]
Hence the foci are
[tex](1,-0.5)\text{ and }(-9,-0.5)[/tex]The equation of the asymptotes are
[tex]y=\frac{b}{a}(x-h)+k\text{ and }y=-\frac{b}{a}(x-h)+k[/tex]Substitute a=4,b=3,h=-4 and k=-1/2, we get
[tex]y=\frac{3}{4}(x-(-4))-\frac{1}{2}\text{ and }y=-\frac{3}{4}(x-(-4))-\frac{1}{2}[/tex][tex]y=\frac{3}{4}(x+4)-\frac{1}{2}\text{ and }y=-\frac{3}{4}(x+4)-\frac{1}{2}[/tex][tex]y=\frac{3}{4}x+\frac{3}{4}\times4-\frac{1}{2}\text{ and }y=-\frac{3}{4}x+(-\frac{3}{4})\times4-\frac{1}{2}[/tex][tex]y=\frac{3}{4}x+3-\frac{1}{2}\text{ and }y=-\frac{3}{4}x-3-\frac{1}{2}[/tex][tex]y=\frac{3}{4}x+3\times\frac{2}{2}-\frac{1}{2}\text{ and }y=-\frac{3}{4}x-3\times\frac{2}{2}-\frac{1}{2}[/tex][tex]y=\frac{3}{4}x+\frac{6-1}{2}\text{ and }y=-\frac{3}{4}x-\frac{(6+1)}{2}[/tex][tex]y=\frac{3}{4}x+\frac{5}{2}\text{ and }y=-\frac{3}{4}x-\frac{7}{2}[/tex][tex]y=0.75x+2.5\text{ and }y=-0.75x-3.5[/tex]Hence the equations of the asymptotes are
[tex]y=0.75x+2.5\text{ and }y=-0.75x-3.5[/tex]Results:
The equation for the hyperbola in standard form is
[tex]\frac{\mleft(x+4\mright)^2}{4^2}-\frac{(y+\frac{1}{2})^2}{3^2}=1[/tex]The vertices are
[tex](0,-0.5)\text{ and }(-8,-0.5)[/tex]The foci are
[tex](1,-0.5)\text{ and }(-9,-0.5)[/tex]The equation of asymptotes are
[tex]y=0.75x+2.5\text{ and }y=-0.75x-3.5[/tex]