# Intersection of two Conics

It is known from algebra that the simultaneous solution set of two equations of the second degree consists of four points. Therefore, two conics will always intersect at four points. These points may all be real and distinct, two real and two imaginary or all imaginary. Two or more points may also coincide.

__Example__**:** Find the points of the intersection of the conics $${x^2} + 4{y^2} = 3$$ and $$2{x^2} – {y^2} = 4$$.

We have two given conics

\[\begin{gathered} {x^2} + 4{y^2} = 3\,\,\,\,{\text{ – – – }}\left( {\text{i}} \right) \\ 2{x^2} – {y^2} = 4\,\,\,\,{\text{ – – – }}\left( {{\text{ii}}} \right) \\ \end{gathered} \]

Now to find the point of intersection of these two conics, solve equations (i) and (ii) by using the method of simultaneous equations.

Multiplying equation (ii) by $$4$$, we get

\[8{x^2} – 4{y^2} = 16\,\,\,\,{\text{ – – – }}\left( {{\text{iii}}} \right)\]

Now adding equation (i) and (iii), we get

\[9{x^2} = 19\,\,\, \Rightarrow x = \pm \frac{{\sqrt {19} }}{3}\]

Putting the value of $$x$$ in equation (ii) to get the variable $$y$$, we have

\[\begin{gathered} 2{\left( { \pm \frac{{\sqrt {19} }}{3}} \right)^2} – {y^2} = 4 \\ \Rightarrow {y^2} = \frac{{38}}{9} – 4 = \frac{{38 – 36}}{9} = \frac{2}{9} \\ \Rightarrow y = \pm \frac{{\sqrt 2 }}{3} \\ \end{gathered} \]

Thus, the points of intersection of the conics are $$\left( { \pm \frac{{\sqrt {19} }}{3}, \pm \frac{{\sqrt 2 }}{3}} \right)$$.