Question

Determine the type of the conic represented by the equation
$x {}^{2} - 2xy + y {}^{2} + 6x - 14y + 29 = 0$
​

Answer 1

Answer:

C

Solution

Comparing the given equation with the equation ax2+2hxy+by2+2gx+2fy+c=0, we have

a=1,b=1,h=−1,c=10,g=10 and f=0  

∴ abc+2fgh−af2−bg2−ch2=10−100=−100≠0.

and, h2−ab=(−1)2−1×1=0⇒h2=ab

Thus, we have

Δ≠0 and h2=ab

So, the given equation represents a parabola.

Step-by-step explanation:

Answer 2

$\large\underline{\sf{Solution-}}$

Given conic is

$\rm \: {x}^{2} - 2xy + {y}^{2} + 6x - 14y + 29 = 0$

On comparing with general equation of conic

$\rm \: {ax}^{2} + 2hxy + {by}^{2} + 2gx + 2fy + c = 0 , \: we \: get$

$a = 1$

$h = - 1$

$b = 1$

$g = 3$

$f = - 7$

$c = 29$

Now, Consider

$\rm \: \triangle = abc + 2fgh - {af}^{2} - {bg}^{2} - {ch}^{2}$

$\rm \: =  \:(1)(1)(29) + 2( - 7)(3)( - 1) - 1 {( - 7)}^{2} - 1 {(3)}^{2} - 29 {( - 1)}^{2}$

$\rm \: =  \:29 + 42 - 49 - 9 - 29$

$\rm \: =  \: - 16$

$\rm\implies \:\triangle \: \ne \: 0$

Now, Consider

$\rm \: ab - {h}^{2}$

$\rm \: =  \:1 \times 1 - {( - 1)}^{2}$

$\rm \: =  \:1 - 1$

$\rm \: =  \:0$

$\rm\implies \:ab - {h}^{2} = 0$

Since,

$\rm\implies \:ab - {h}^{2} = 0 \: \: and \: \: \triangle \: \ne \: 0$

$\rm\implies \:{x}^{2} - 2xy + {y}^{2} + 6x - 14y + 29 = 0 \: is \: parabola.$

$\rule{190pt}{2pt}$

Additional Information :-

Let us consider general equation of conic

$\rm \: {ax}^{2} + 2hxy + {by}^{2} + 2gx + 2fy + c = 0$

$\\ \begin{gathered}\boxed{\begin{array}{c|c} \bf Condition & \bf Nature \: of \: conic \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf \triangle \ne \: 0, \: ab - {h}^{2} > 0 & \sf Ellipse \\ \\ \sf \triangle \ne \: 0, \: ab - {h}^{2} < 0 & \sf Hyperbola \\ \\ \sf \triangle \ne \: 0, \: ab - {h}^{2} > 0, \: a + b = 0 & \sf Hyperbola(rectangular)\\ \\ \sf \triangle \ne \: 0, \: h = 0, \: a = b & \sf Circle \end{array}} \\ \end{gathered}$