Conic Section

A Point, a Line and a Parabola

A Point and a Parabola

Let $P(x_1,y_1)$ be any point and $y^2=4ax$ be a parabola. Draw $PM$ perpendicular to x-axis. Let $PM$ meet the parabola at $Q$.

A point and a parabola

Then $P$ lies outside, on or inside the parabola according as $PM$ $>$, $=$ or $<$ $QM$. Here, \[PM=y_1\] \[\text{and, }\;QM\:\text{ at }\:x_1=\sqrt{4ax_1}\]

Hence, $P$ lies outside the parabola if \[PM>QM\] \[PM^2>QM^2\] \[\therefore y_1^2>4ax_1\] $P$ lies on the parabola if \[PM=QM\] \[PM^2=QM^2\] \[\therefore y_1^2=4ax_1\] $P$ lies inside the parabola if \[PM<QM\] \[PM^2<QM^2\] \[\therefore y_1^2<4ax_1\]

A Line and a Parabola

Let the equation of a parabola be \[y^2=4ax\] and, the equation of a line be \[y=mx+c\] Now, let us solve these two equations simultaneously to obtain the points of intersection of the parabola and the line. For this, eliminating $y$, \[(mx+c)^2=4ax\] \[m^2x^2+2mcx+c^2-4ax=0\] \[m^2x^2+2(mc-2a)x+c^2=0\text{ __(1)}\] This is quadratic in $x$. Hence, $x$ has two values and corresponding to the two values of $x$, we can obtain the two values of $y$. In this way, we can obtain the points of intersection.

Now, the determinant of equation $\text{(1)}$ is \[B^2-4AC\] \[4(mc-2a)^2-4m^2c^2\] \[=4(m^2c^2-4mca+4a^2-m^2c^2)\] \[=16a(a-mc)\]

  • If $a-mc>0$ or $c<\frac{a}{m}$, then the roots of $\text{(1)}$ i.e. the values of $x$ are real and distinct. In this case, the line cuts the parabola at two distinct points.
  • If $a-mc=0$ or $c=\frac{a}{m}$, then the roots of $\text{(1)}$ are real and equal. In this case the line meets the parabola at two coincident points. We say that the line is tangent to the parabola.
  • If $a-mc<0$ or $c>\frac{a}{m}$, then the roots of $\text{(1)}$ are imaginary. In this case, the line does not intersect the parabola. But it is customary to say that the line cuts the parabola at the imaginary points.

Hence, the line will touch the parabola if it intersects the parabola in two coincident points i.e. if the roots are real and equal. \[\therefore B^2-4AC=0\] \[4(mc-2a)^2-4m^2c^2=0\] \[16a(a-mc)=0\] \[\therefore c=\frac{a}{m}\] This is the condition of tangency. Thus, the equation of a line tangent to the parabola is given by, \[y=mx+\frac{a}{m}\] This equation of the tangent is in slope form.