# Find the equation of the parabola given vertex and directrix?

The vertex of parabola is $$V(3, 1)$$ and the directrix is $$4x + 3y = 5$$.

I can't figure out the focus by using the axis of symmetry which is $$3x - 4y -5 =0$$.

• Do you need to find the parabola’s focus or its equation (per the title)? You can do the latter without the focus.
– amd
Oct 25, 2019 at 7:01
• Sorry for the late comment but I am really curious about your method.Can you please give me some hint? Oct 27, 2019 at 3:43

## 3 Answers

Here is a straightforward way to find the unique focal point. From the directrix and the symmetry lines $$4x+3y=5, \>\>\>\>\>3x-4y=5$$

their intersection is $$(\frac75, -\frac15)$$. Since the vertex is the midpoint between the focus $$(a,b)$$ and the intersection point,

$$3=\frac{a+\frac75}{2},\>\>\>\>\>1=\frac{b-\frac15}{2}$$

Then, solve for the focus $$(a, b)$$.

• Dear Senor/Senora, Can you tell me please why didn't I come up with this simple thing?I have been working on this problem for hours. Oct 24, 2019 at 20:48
• @Ghost - I often find it very useful to visualize the geometric relationship among various components in question and try to discovery their linkage. Oct 24, 2019 at 21:06

You can't use the directrix as axis of symmetry since the vertex and the focus of the parabola are on the same half-plane in relation to the directrix. However, you can use the point-line distance formula to find the focal length $$d$$ of the parabola. That is

$$d=\frac{|4(3)+3(1)-5|}{\sqrt{4^2+3^2}}=2$$

Then, you can also find the focal axis of the parabola by finding the equation of the perpendicular line to the directrix that passes through $$(3,1)$$, which is $$y-1=\frac{3}{4} (x-3)$$ $$3x-4y-5=0$$

Now, the focus will be on a point $$(p,q)$$ that satisfies $$3p-4q-5=0$$ and $$(p-3)^2+(q-1)^2=4$$. One of the solutions to the previous system is $$(\frac{23}{5}, \frac{11}{5})$$, which is the focus of the parabola.

This algebraic process has a geometric equivalent: Take the point $$A(3,1)$$ as the center of a circumference with radius $$d=2$$, this circumference intersects the line $$3x-4y-5=0$$ in 2 points $$B\big(\frac{7}{5}, -\frac{1}{5}\Big)$$ and $$C\Big(\frac{23}{5}, \frac{11}{5}\Big)$$, as seen in the image below.

Both of those points are on the focal axis and also are at a distance of 2 units from the vertex. Visually, only C must be focus, but algebraically, you would have to check which of those points do not belong to the directrix. In this case, for the coordenates of point $$B$$ you have $$3\Big(\frac{7}{5}\Big)-4\Big(-\frac{1}{5}\Big)-5=\frac{21}{5}+\frac{4}{5}-\frac{25}{5}=0$$ Since the coordinates of point $$B$$ satisfy the equation $$3x-4y-5=0$$ this means $$B$$ lies on the directrix, and therefore, $$C$$ must be the focus.

• The thing is Focus is (23/5 , 11/5) please correct your answer.But There is another problem that I don't understand.Why I ended up with two focus.I mean there should be one focus right?Can you please explain? Oct 24, 2019 at 20:30
• Sorry, I corrected my answer. Also I added an edit to explain further why only one of the solutions is the focus. Oct 25, 2019 at 18:11

The parabola’s focus is easily found via, say, a vector computation: The vertex is midway between the focus and directrix. The signed distance from the directrix to the vertex is $${4\cdot3+3\cdot1-5\over5}=2$$ and from the equation of the directrix the corresponding unit normal is $$\frac15(4,3)$$, so the focus is at $$(3,1)+\frac25(4,3)=\left(\frac{23}5,\frac{11}5\right)$$. However, as I mentioned in a comment, you can find an equation for the parabola without explicitly computing the focus.

Via a combination of rotation and translation, any parabola’s Cartesian equation can be brought into the canonical form $$Y^2=4pX$$, where $$p$$ is the focal distance. If you apply a rotation and translation to this equation, you’ll get something of the form $$(ax+by+c)^2=4p(bx-ay+d)$$, with $$a^2+b^2=1$$. The equation $$ax+by+c=0$$ gives the symmetry axis, while $$bx-ay+d=0$$ is the tangent at the vertex. You can see why this might be so by comparing the transformed equation to the canonical form: The $$X$$-axis, with equation $$Y=0$$ is the axis of symmetry, and it is transformed to the line $$ax+by+c=0$$. Similarly, the $$Y$$-axis with equation $$X=0$$ is the tangent at the vertex, and that is mapped to $$bx-ay+d=0$$.

The tangent at the vertex is parallel to the directrix, so its equation can be found via the point-normal form: $$4x+3y=4\cdot3+3\cdot1=15$$. The corresponding equation for the axis of symmetry is therefore $$3x-4y=3\cdot3-4\cdot1=5$$. Normalizing both equations by dividing by $$\sqrt{4^2+3^2}=5$$ and using the focal distance computed at the top, we get for an equation of the parabola $$\frac1{25}\left(3x-4y-5\right)^2=\frac85\left(4x+3y-15\right).$$ Rearrange and simplify as needed.