conic sectionscoordinate systemsgeometry
Suppose the equation of latus rectum is x=4 and the vertex is (2,3). I am confused wouldn't there be many parabola with this same vertex and latus rectum.If not how to find the equation?
The answer that is given is $$(y-3)^2=8(x-2)$$ in the textbook. I am not getting it. Thanks in advance.
Let $P(7,13),F(-1,-1)$. Also, let $T$ be the intersection point of the line $y=3x-8$ with the axis of symmetry. Let $V$ be the vertex, and let $K$ be the point on the axis of symmetry such that $PK$ is perpendicular to the axis.
We use the followings (for the proof, see the end of this answer) :
(1) $PF=TF$
(2) $VT=VK$
(3) $\text{(the length of the latus rectum)}=4\times FV$
First of all, setting $T$ as $(t,3t-8)$ where $t\not= 7$ and using $(1)$ give $$(-1-7)^2+(-1-13)^2=(-1-t)^2+(-1-3t+8)^2\quad\Rightarrow\quad t=-3\quad\Rightarrow\quad T(-3,-17)$$
Hence, the axis of symmetry is the line $TF$ : $y=8x+7$. So, the line $PK$ is $y-13=(-1/8)(x-7)$, i.e. $y=-x/8+111/8$ from which $K(11/13,179/13)$ follows.
From $(2)$, since $V$ is the midpoint of the line segment $TK$, we have $V(-14/13,-21/13).$
Finally, using $(3)$, we get that the answer is $\color{red}{4\sqrt{5/13}}$.
Proof for $(1)(2)(3)$ :
We may suppose that the equation of a parabola is $y^2=4px$ where $p\gt 0$.
$\qquad\qquad\qquad$
We consider the tangent line at $A(a,b)$ where $b^2=4pa$ with $b\gt 0$. Let $B$ be the intersection point of the tangent line with $x$ axis which is the axis of symmetry. Also, let $C(p,0)$ be the focus, and let $D(a,0)$ be a point on $x$ axis such that $AD$ is perpendicular to $x$ axis. The vertex is $O(0,0)$, and let $E(p,e)$ where $e\gt 0$ be the intersection point of the parabola with the line perpendicular to $x$ axis passing through $C$.
(1)
Since the equation of the tangent line at $A$ is given by $by=2p(x+a)$, we have $B(-a,0)$, and so $$AC=\sqrt{(a-p)^2+(b-0)^2}=\sqrt{a^2-2ap+p^2+4pa}=\sqrt{(p+a)^2}=p+a=BC.$$
(2)
$OB=0-(-a)=a=OD$.
(3)
Solving $y^2=4px$ and $x=p$ gives $y=\pm 2p$, and so $e=2p$. Hence, $$\text{(the length of the latus rectum )}=2\times EC=2e=4p=4\times OC.$$
You’ve done some nice work in finding the Cartesian equation for the parabola, but it think it might be easier overall to work with the parametric form $\mathbf r:t\mapsto(x(t),y(t))$. (I’ve omitted most of the tedious details of the algebraic manipulations in the following.)
The first order of business is to find the angle of the parabola’s axis of symmetry. One way to do this is to use the fact that for any two points on a parabola, the line defined by their midpoint and the intersection of the tangents at the two points is parallel to this axis. Taking $t=\pm1$ is reasonably convenient. The midpoint is simply $(\mathbf r(1)+\mathbf r(-1))/2$. For the intersection of the tangents, we have $\mathbf r(1)+s\mathbf r'(1)=\mathbf r(-1)+t\mathbf r'(-1)$, which expands into the system $$\begin{align}s(2a+b)+a+b+c &= t(-2a+b)+a-b+c \\ s(2a'+b')+a'+b'+c'&=t(-2a'+b')+a'-b'+c\end{align}$$ with solution $s=-1$, $t=1$. The direction vector for the parabola’s axis is thus $$\begin{align}\frac12(\mathbf r(1)+\mathbf r(-1))-(\mathbf r(-1)+\mathbf r'(-1)) &= \frac12(\mathbf r(1)-\mathbf r(-1))-\mathbf r'(-1) \\ &=(b,b')-(b-2a,b'-2a') \\ &=(2a,2a'),\end{align}$$ so we can take $\mathbf n=(a,a')$ as the direction of the axis.
Next, we find the parabola’s vertex. The tangent to the parabola at the vertex is orthogonal to its axis, which gives rise to the equation $$\mathbf n\cdot\mathbf r'(t)=(a,a')\cdot(at+b,a't+b')=a(at+b)+a'(a't+b')=0.$$ Solving for $t$ we get $$t_c=-\frac12{ab+a'b'\over a^2+a'^2}\tag{1}.$$ You can verify that the function $\mathbf r$ is symmetric with respect to this point in the sense that the chord defined by the points $\mathbf r(t_c\pm\Delta t)$ is orthogonal to the parabola’s axis, so the two points are equidistant from the vertex.
The tangents at the ends of the latus rectum meet the axis at a 45° angle, which means that they are orthogonal to each other. This fact leads to the equation $$\mathbf r'(t_c+\Delta t)\cdot\mathbf r'(t_c-\Delta t)=(2a(t_c+\Delta t)+b)(2a(t_c-\Delta t)+b)+(2a'(t_c+\Delta t)+b')(2a'(t_c-\Delta t)+b')=0$$ for the ends of the latus rectum, which has the solutions $$\Delta t=\pm\frac12{ab'-a'b\over a^2+a'^2}\tag{2}.$$ Because of the symmetry noted previously, we know that both of these solutions represent the same pair of points.
Finally, we compute the length of the latus rectum: $$\begin{align}\|\mathbf r(t_c+\Delta t)-\mathbf r(t_c-\Delta t)\|^2 &= (\mathbf r(t_c+\Delta t)-\mathbf r(t_c-\Delta t))\cdot(\mathbf r(t_c+\Delta t)-\mathbf r(t_c-\Delta t)) \\ &= \left({a'(ab'-a'b)^2\over(a^2+a'^2)^2}\right)^2+\left({a(ab'-a'b)^2\over(a^2+a'^2)^2}\right)^2 \\ &={(ab'-a'b)^4\over(a^2+a'^2)^3},\end{align}$$ so the length of the latus rectum is $${(ab'-a'b)^2\over(a^2+a'^2)^{3/2}}.\tag{3}$$ With this distance in hand, you can now easily find the parabola’s focus and directrix, if needed.
Incidentally, this is another path to a Cartesian equation for this parabola. If you have the directrix given by an equation in the form $\mathbf n\cdot\mathbf x=d$ and the focus $\mathbf f$, then an equation of the parabola is $$\left({d-\mathbf n\cdot\mathbf x\over\|\mathbf n\|}\right)^2=(\mathbf x-\mathbf f)\cdot(\mathbf x-\mathbf f).$$
To continue working with the Cartesian equation instead, I’d take a slightly different approach than in the cited paper. First look at the equation of a parabola with axis parallel to the $y$-axis, $y=ax^2+bx+c$. For such a parabola, the length of the latus rectum is simply $|1/a|$. For the general parabola $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$, we take Erick Wong’s suggestion to rotate so as to eliminate the quadratic terms involving $y$. The equation will then be of the form $A'x'^2+D'x'+E'y'+F=0$ (note that the constant term is unchanged by a rotation), with latus rectum length $|E'/A'|$. Invariance of the trace tells us that $A'=A+C$, but finding $E'$ will take a a bit more work.
For a parabola, $B^2=4AC$, so we can rewrite the general equation as $(\alpha x+\beta y)^2+Dx+Ey+F=0$, first multiplying through by $-1$ if necessary to make $A$ and $C$ positive. In matrix form this is $\begin{bmatrix}x&y&1\end{bmatrix}M\begin{bmatrix}x&y&1\end{bmatrix}^T=0$, with $$M=\begin{bmatrix}\alpha^2&\alpha\beta&D/2\\\alpha\beta&\beta^2&E/2\\D/2&E/2&F\end{bmatrix}.$$ The direction of the parabola’s axis is given by an eigenvector of $0$ (i.e., an element of the kernel) of the quadratic part of this matrix, one of which is $\begin{bmatrix}-\beta&\alpha\end{bmatrix}^T$. We want to rotate so as to bring this vector parallel to the $y$-axis. The appropriate rotation is $$R=\begin{bmatrix}{\alpha\over\sqrt{\alpha^2+\beta^2}}&-{\beta\over\sqrt{\alpha^2+\beta^2}}&0\\{\beta\over\sqrt{\alpha^2+\beta^2}}&{\alpha\over\sqrt{\alpha^2+\beta^2}}&0\\0&0&1\end{bmatrix}$$ and $$R^TMR = \begin{bmatrix}\alpha^2+\beta^2&0&\frac12{\alpha D+\beta E\over\sqrt{\alpha^2+\beta^2}} \\ 0&0&\frac12{\alpha E-\beta D\over\sqrt{\alpha^2+\beta^2}} \\ \frac12{\alpha D+\beta E\over\sqrt{\alpha^2+\beta^2}} & \frac12{\alpha E-\beta D\over\sqrt{\alpha^2+\beta^2}} & F\end{bmatrix}.$$ Comparing this to the result in the first paragraph above, we find that the latus rectum length for the general equation is $${|\alpha E-\beta D|\over(\alpha^2+\beta^2)^{3/2}} = {|E\sqrt A\mp D\sqrt C|\over(A+C)^{3/2}}.$$ Choose the sign opposite to that of $B$. Many of the terms in the Cartesian equation you derived are constant, so pulling the coefficients of $x$ and $y$ out of it to construct $D$ and $E$ doesn’t look too bad.
Best Answer
Let $BE$ be the directrix
As the axis of the parabola is perpendicular to the latus rectum, the equation of the axis(OA) $y=3$
So, $c=a=3$
As the vertex O$(2,3)$ is the mid-point of $A,B, \frac{b+4}2=2\implies b=0$
So, the equation of the directrix(BE) will be $x=0$ as it is perpendicular to the axis.
If P$(h,k)$ be any point on the parabola,
the distance of $AP=\sqrt{(h-4)^2+(k-3)^2}$
the distance of P$(h,k)$ from the directrix $BE$ will be $|h|$
As the eccentricity of a parabola is $1$
So, $$\sqrt{(h-4)^2+(k-3)^2}=|h|$$
Squaring we get $(h-4)^2+(k-3)^2=h^2\implies (k-3)^2=h^2-(h-4)^2=8h-16=8(h-2)$
So, the equation of the parabola : $(y-3)^2=8(x-2)$