0
$\begingroup$

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 $3x - 4y -5 =0$.

$\endgroup$
2
  • $\begingroup$ Do you need to find the parabola’s focus or its equation (per the title)? You can do the latter without the focus. $\endgroup$
    – amd
    Oct 25, 2019 at 7:01
  • $\begingroup$ Sorry for the late comment but I am really curious about your method.Can you please give me some hint? $\endgroup$
    – Ghost
    Oct 27, 2019 at 3:43

3 Answers 3

1
$\begingroup$

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 $(a, b)$.

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$
    – Ghost
    Oct 24, 2019 at 20:48
  • $\begingroup$ @Ghost - I often find it very useful to visualize the geometric relationship among various components in question and try to discovery their linkage. $\endgroup$
    – Quanto
    Oct 24, 2019 at 21:06
1
$\begingroup$

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. enter image description here

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.

$\endgroup$
2
  • $\begingroup$ 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? $\endgroup$
    – Ghost
    Oct 24, 2019 at 20:30
  • $\begingroup$ Sorry, I corrected my answer. Also I added an edit to explain further why only one of the solutions is the focus. $\endgroup$ Oct 25, 2019 at 18:11
0
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .