conic sectionscoordinate systemsgeometry
The curve passing through the points of intersection of $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $x^2+y^2+2gx+2fy+c=0$ represents a pair of straight lines which are:
(a) equally inclined to the x axis
(b) perpendicular to each other
(c) pass through a fixed point
(d) none of the above
I am having a problem comprehending the question. How can the 2 intersection points define a whole curve? How can 2 distinct lines be defined by 2 intersection points?
A function $f:\mathbb{R}^n \to \mathbb{R}$ is homogeneous of degree $r$ when it has the following property:
$$f(\lambda x_1,\lambda x_2,\ldots,\lambda x_n) = \lambda^r f(x_1,x_2,\ldots,x_n)$$
Examples of such functions include:
Linear functions, they are of degree 1. If you scale the graph of the function by a factor $\lambda$, you still get the same graph, except that all points have coordinates scaled up by the factor $\lambda$.
The monomials in one variable: $f:\mathbb{R} \to \mathbb{R}: x \mapsto x^n$ is homogeneous of degree $n$, scale $x$ by a factor $\lambda$ and the function will scale by a factor $\lambda^n$.
Any polynomial (function) in $n$ variables such that each term appearing in the polynomial is of degree $k$ is homogeneous of degree $k$. For instance $x^2 y + z^3$ is homogeneous of degree $3$.
The importance of homogeneity is the scale invariance of the functions. Which implies that the graphs of the functions will be scale invariant. Indeed, imagine a homogeneous function is used to define a geometrical object implicitly:
$$f(x_1,x_2,\ldots,x_n)=0$$
meaning all points with coordinates $(x_1,\ldots,x_n)$ that satisfy this equation will belong to the geometrical figure defined by $f$. If $f$ is homogeneous, it immediately follows that any multiple of these coordinates also satisfies the equations. In other words, any point that satisfies the equation immediately implies the entire ray going through that point and the origin of the space belong to the geometrical object.
Homogenizing an implicit polynomial equation means adding an extra variable $z$ and multiply any term by $z^k$ with $k$ such that the resulting polynomial is homogeneous. Of course, since any $z$-multiple of the polynomial will also be homogeneous, you choose the resulting homogeneous polynomial with smallest possible degree.
In your example, this would become
$$ax^2 + 2hxy + by^2 + 2gxz + 2fyz + cz^2 = 0$$
If you take the intersection of the geometrical figure defined by above equation with the plane $z=1$, you get back the original figure. If you choose another plane parallel with the $z=1$ plane, you get a scaled up or scaled down version of the figure.
@amd's answer gave me the hint.
First, if we take $\lambda$ to be $-1$ then we get a non homogeneous equation of first degree($x+y=4$) which would surely represent a straight line passing through the points of intersection of the two curves. Now if we somehow homogenize the equation of the second circle withe the help of the equation of the line, i.e $x+y=4$, to a second degree equation, then this equation would surely represent a pair of straight lines through the origin, passing through the points of intersection of the two curves.
I did this by writing $1=\frac{x+y}{4}$ and so multiplying this thing in the RHS with all the terms of first degree in the equation of the second curve and the square of this to the term of zeroth degree, i.e the constant $+14$.
This gives the equation of the pair of straight lines:
$$x^2+y^2-6x\left(\frac{x+y}{4}\right)-6y\left(\frac{x+y}{4}\right)+14\left(\frac{x+y}{4}\right)^2=0$$
Best Answer
Hint
$$x^2+y^2+2gx+2fy+c=0\to (x+g)^2+(y+f)^2=g^2+f^2-c$$
which can be a circle.
That is a picture about what is happening according to the values of $a,b,c,f,g$.
Can you go forward?