Find the equation of the parabola whose directrix is…

Find the equation of the parabola whose directrix is $2x+y-1=0$ and which is tangent to the line $x-y=2$ at the point $M_0(0,-2)$.

Step 1: Locating the focus (geometric method)

Reflection property: the reflection $F^*$ of the focus in any tangent is the foot of the perpendicular from the point of contact to the directrix. (The tangent bisects the angle between $M_0F$ and the perpendicular $M_0F^*$, and $|M_0F|=|M_0F^*|$.)

Foot of $M_0$ on the directrix. From $2\cdot0+(-2)-1=-3$ we get $t=-\tfrac35$, so

\[ F^*=M_0+\tfrac35(2,1)=\left(\tfrac65,-\tfrac75\right). \]

Reflect $F^*$ in $x-y-2=0$. Here $\tfrac65+\tfrac75-2=\tfrac35$, so

\[ F=\left(\tfrac65-\tfrac35,\ -\tfrac75+\tfrac35\right)=\left(\tfrac35,-\tfrac45\right). \]

Step 2: The source's analytic method

With $F=(p,q)$, the source sets up three conditions:

• $M_0$ lies on the parabola: $p^2+(q+2)^2=\tfrac95$.
• The tangent at $M_0$ has slope $1$ (implicit differentiation): $p+q=-\tfrac15$.
• The reflection of $F$ in the tangent lies on the directrix: $p+2q+1=0$.

The two linear equations give $p=\tfrac35$, $q=-\tfrac45$, and the first is then satisfied: $\tfrac9{25}+\tfrac{36}{25}=\tfrac95$. $\checkmark$

Step 3: Equation

\[ \Big(x-\tfrac35\Big)^2+\Big(y+\tfrac45\Big)^2=\frac{(2x+y-1)^2}{5} \]

\[ 5x^2+5y^2-6x+8y+5=4x^2+y^2+4xy-4x-2y+1 \]

\[x^2-4xy+4y^2-2x+10y+4=0\quad\Longleftrightarrow\quad(x-2y)^2=2x-10y-4\]

Step 4: Verification

• $M_0$: $0-0+16-0-20+4=0$. $\checkmark$
• Slope at $M_0$: $2x-4y-2+(-4x+8y+10)y'=0$ gives $6-6y'=0$, so $y'=1$. $\checkmark$
• Type: $B^2-4AC=16-16=0$, a parabola. Its axis is parallel to $(2,1)$, perpendicular to the directrix. The parameter is $p=d(F,\text{directrix})=\tfrac{3}{5\sqrt5}$, and the vertex is $\left(\tfrac{18}{25},-\tfrac{37}{50}\right)$.