The point  (-11,8)

It cannot be a conic...it has to be a point.

If sum of 2 non-negative quantities is zero then both the quantities have to be zero.

Solving the 2 linear equations obtained on equating them to zero gives the answer.

(3x+4y+1)^2+(x+y+3)^2=0

On expanding we get,

4x.x + 5y.y + 13xy + 7y + 6x + 10=0

it is in the form - ax.x + by.y + 2hxy + 2gx + 2fy + c=0.

to check we first find the area i.e.   abc + 2fgh - af.f - bg.g - ch.h= area.

it is non zero,  hence it does not represent the straight lines.

now, we check, h.h= 169/4 and a.b= 20.

here, h.h>a.b ;  hence, the equation represents the hyperbola with center  (-11, 8).

Guys!! Doesn't that equation just represent a point ?!! The point of intersection of those 2 linear equations..

If p(x) and q(x) are 2 linear expressions in x of degree 1,

and if

then since square of anything real can never be negative, both have to be zero.

Apply this sense in the above question...

The corrected version

3x+4y+1)^2+(x+y+3)^2=0

On expanding we get,

10x.x + 17y.y + 26xy + 14y + 12x + 10=0

it is in the form - ax.x + by.y + 2hxy + 2gx + 2fy + c=0.

to check we first find the area i.e.   abc + 2fgh - af.f - bg.g - ch.h= area.

it is non zero,  hence it does not represent the straight lines.

now, we check, h.h= 169 and a.b= 170.

here, h.h<a.b ;  hence, the equation represents the ellipse with center  (-11, 8).

If it is a point that means if we check the radius that should come out to be ZERO. But, sqrt(g^2 + f^2 - c) is not equal to 0.  Therefore, the equation does not represent the point.  (where g=6, f=7, and c=10 )

It is time to force matters...

Let a,b be any 2 REAL numbers.

The facts are

So if I say

there can be ONLY ONE solution.

Now if a,b are functions of x,y returning real values, then both the functions a and b must be zero as shown above.

Let a(x,y)=(3x+4y+1)

and b(x,y)=(x+y+3)

So the ONLY solutions to

are

Using this we put a(x,y)=0 and b(x,y)=0 . What we get now is

3x+4y+1=0 AND x+y+3=0

These are a pair of simultaneous linear equations which can only represent ONE point..

If anybody argues that it is a conic, I am ready to accept that fact if you can find just one more point which satisfies that equation.

And Deevita, if you call it an ellipse with centre (-11,8) can you justify (-11,8) satisfying the given equation?

An ellipse which passes through its centre?? LOL....it's ok...we all make mistakes, right?

Just one more thing to support my point,

when the equation is written in the general equation of 2nd degree form, then

abc+2fgh-a.f.f-b.g.g-c.h.h comes out to be zero.

and h.h < a.b

which proves that the equation represents a pair of imaginary lines intersecting in a real point.

Your argument is appropriate and satisfactory.  I was mistaken. Now, it seems proper to me.