Question

show that the equation of second degree 5x2-2xy+5y2+2x-10y-7=0

Answer 1

Answer:

Step-by-step explanation:

We know that the general equation of second degree

$ax^2+2hxy+by^2+2gx+2fy+c=0\text{.............................equation-1}$

Given equation

$5x^2-2xy+5y^2+2x-10y-7=0\\\\\Rightarrow 5x^2+(-2)xy+5y^2+2x+(-10)y+(-7)=0\text{...............................equation-2}$

Now comparing the two equation we get,

$a=5$

$b=5$

$c=-7$

$2f=-10\\\\\therefore f=-5$

$2h=-2\\\\\therefore h=-1$

$2g=2\\\\\therefore g=1$

Answer 2

Answer:

According to the equation of conic, the given equation represents ellipse.  

Step-by-step explanation:

Given equation is 5x^2 - 2xy + 5y^2 + 2x - 10y - 7 = 0  ..........(1)

we know that the general equation of Conic is  

          [ ( Ax^2 ) + 2hxy + ( By^2 ) + 2gx + 2fy + C ] = 0 ..........(2)

Compare equation equation 1 and equation 2 ,

we get, A = 5,    

B = 5,  

g = 1,  

f = -5,  

C = -7,  and

h = -1

General formula of Δ ( area ) is [ ABC + 2fgh - Af² - Bg² - Ch² ]  

Now have to put the value of a, b, c, g, h, f in above equation    

[ -175 +10 -125 -5 +7 ] = -288

Δ = -288

now you have to calculate AB - h² = 25 - 1 = 24

Therefore, the given equation represents ellipse.