Question

Can you please help me with 24Write the equation for the hyperbola in standard form if it is not already, and identify the vertices and foci, and write equation of asymptotes

Answer 1

24)

Given:

`-9x^2+72x+16y^2+16y+4=0`

We need to find the hyperbola in standard form

`((x-h)^2)/(a^2)-((y-k)^2)/(b^2)=1`

The given equation can be written as follows.

`-9(x^2+8x)+16(y^2+y)+4=0`

`-9(x^2+2*4x+4^2-4^2)+16(y^2+2*(1)/(2)y+(1)/(2^2)-(1)/(2^2))+4=0`

`-9(x^2+2*4x+4^2)-(-9)4^2+16(y^2+2*(1)/(2)y+(1)/(2^2))-(16)(1)/(2^2)+4=0`

`-9(x^2+2*4x+4^2)+144+16(y^2+2*(1)/(2)y+(1)/(2^2))-4+4=0`

`-9(x^2+2*4x+4^2)+144+16(y^2+2*(1)/(2)y+(1)/(2^2))=0`
`\text{ Use (a+b)}^2=a^2+2ab+b^2\text{.}`

`-9(x+4)^2+144+16(y+(1)/(2))^2=0`

Subtracting 144 from both sides, we get

`-9(x+4)^2+144+16(y+(1)/(2))^2-144=0-144`

`-9(x+4)^2+16(y+(1)/(2))^2=-144`

Dividing both sides by (-144), we get

`(-9\mleft(x+4\mright)^2)/(-144)+(16(y+(1)/(2))^2)/(-144)=-(144)/(-144)`

`(\mleft(x+4\mright)^2)/(16)-((y+(1)/(2))^2)/(9)=1`

`(\mleft(x+4\mright)^2)/(4^2)-((y+(1)/(2))^2)/(3^2)=1`

Hence we get the standard form of the hyperbola.

`((x-h)^2)/(a^2)-((y-k)^2)/(b^2)=1`

Comapring with standerd form, we get

`h=-4,k=-(1)/(2),a=4\text{ and b=3.}`

The coordinates of the vertices are

`(h+a,k)\text{ and }(h-a,k)\text{ }`

Substitute h=-4, a=4 and k=-1/2.

`(-4+4,-(1)/(2))\text{ and }(-4-4,-(1)/(2))`

`(0,-(1)/(2))\text{ and }(-8,-(1)/(2))`

`(0,-0.5)\text{ and }(-8,-0.5)`

Hence the vertices are (0, -0.5) and (-8, -0.5).

The distance between foci is 2c,

`c^2=a^2+b^2`

Substitute a=4 and b=3 to find the value of c.

`c^2=4^2+3^2`

`c^2=16+9=25^{}`
`c=\pm\sqrt[]{25}`
`c=\pm5`

The measure of distance can not be negative.

`c=5`

The coordinates of the foci are

`(h+c,k)\text{ and }(h-c,k)`

Substitute h=-4, k=-1/2 and c=5, we get

`(-4+5,-(1)/(2))\text{ and }(-4-5,-(1)/(2))`

`(1,-(1)/(2))\text{ and }(-9,-(1)/(2))`
`(1,-0.5)\text{ and }(-9,-0.5)`

Hence the foci are

`(1,-0.5)\text{ and }(-9,-0.5)`

The equation of the asymptotes are

`y=(b)/(a)(x-h)+k\text{ and }y=-(b)/(a)(x-h)+k`

Substitute a=4,b=3,h=-4 and k=-1/2, we get

`y=(3)/(4)(x-(-4))-(1)/(2)\text{ and }y=-(3)/(4)(x-(-4))-(1)/(2)`

`y=(3)/(4)(x+4)-(1)/(2)\text{ and }y=-(3)/(4)(x+4)-(1)/(2)`

`y=(3)/(4)x+(3)/(4)*4-(1)/(2)\text{ and }y=-(3)/(4)x+(-(3)/(4))*4-(1)/(2)`

`y=(3)/(4)x+3-(1)/(2)\text{ and }y=-(3)/(4)x-3-(1)/(2)`

`y=(3)/(4)x+3*(2)/(2)-(1)/(2)\text{ and }y=-(3)/(4)x-3*(2)/(2)-(1)/(2)`

`y=(3)/(4)x+(6-1)/(2)\text{ and }y=-(3)/(4)x-((6+1))/(2)`

`y=(3)/(4)x+(5)/(2)\text{ and }y=-(3)/(4)x-(7)/(2)`

`y=0.75x+2.5\text{ and }y=-0.75x-3.5`

Hence the equations of the asymptotes are

`y=0.75x+2.5\text{ and }y=-0.75x-3.5`

Results:

The equation for the hyperbola in standard form is

`(\mleft(x+4\mright)^2)/(4^2)-((y+(1)/(2))^2)/(3^2)=1`

The vertices are

`(0,-0.5)\text{ and }(-8,-0.5)`

The foci are

`(1,-0.5)\text{ and }(-9,-0.5)`

The equation of asymptotes are

`y=0.75x+2.5\text{ and }y=-0.75x-3.5`