Analytic Geometry – Problem 5: Find the equation of parabola

For parabola $E:y^2=2px\,(p>0)$, the directrix is $x=-1$. Line $l:x=my+3$ intersects the parabola at points $A,B$.

(1) Find the equation of parabola $E$.

(2) If $AB=4\sqrt{15}$, find $m$.

(3) If there exist two points $C,D$ on parabola $E$ that are symmetric with respect to line $l$, find the range of $m$.

(1) Since the directrix is $x=-\dfrac p2=-1$, we get $p=2$. Therefore $$E:y^2=4x.$$

(2) Solve $$\begin{cases} y^2=4x,\\ x=my+3 \end{cases} \Rightarrow y^2-4my-12=0.$$ Let its roots be $y_1,y_2$. Then $$y_1+y_2=4m,\quad y_1y_2=-12.$$ The chord-length formula gives $$AB=\sqrt{1+m^2}\,|y_1-y_2|=\sqrt{1+m^2}\sqrt{16m^2+48}.$$ From $AB=4\sqrt{15}$, $$m^4+4m^2-12=0\Rightarrow m=\pm\sqrt2.$$

(3) Let $$C\left(\frac{a^2}4,a\right),\quad D\left(\frac{b^2}4,b\right).$$ The case $m=0$ does not satisfy the symmetry requirement. For $m\ne0$, from $CD\perp l$, $$a+b=-\frac4m.$$ The midpoint of $CD$ lies on $l$, and we obtain that its $x$-coordinate equals 1, hence $$\frac{a^2+b^2}8=1\Rightarrow a^2+b^2=8.$$ Combining with $a+b=-\frac4m$, the solvability condition of the resulting quadratic in $b$ is $$\Delta>0\Rightarrow m^2>1.$$ So $$m>1\quad\text{or}\quad m<-1.$$

(1) The parabola is $$y^2=4x.$$ This is found from the directrix relation $x=-\frac p2$.

(2) Using the intersection quadratic and the chord-length formula, $m$ satisfies $m^4+4m^2-12=0$, so $$m=\pm\sqrt2.$$

(3) Imposing the symmetry constraints (perpendicular chord and midpoint on $l$) leads to $m^2>1$. Therefore the range is $$m>1\ \text{or}\ m<-1.$$