analytic geometryconic sections
$$y = x \tan \theta – \frac{gx^2}{2u^2 \cos^2 \theta}$$
Find the focus and length of the latus rectum of this parabola.
Here $u$ is a constant. We also know tangent at the origin makes an angle ($\theta$) with the $x$-axis.
I know the focus and length of the latus rectum of $x^2 = 4ay$, but this is not of that form: the parabola is rotated. How to handle this case?
First simplify the parabola equation to a general $2^{nd}$ degree equation: $$(\alpha x+\beta y)^2+(2\beta\gamma - B)y+(2\alpha\gamma-A)x+\gamma^2-C=0$$
Now let $2g=2\alpha\gamma-A$,$2f=2\beta\gamma - B$ and $d=\gamma^2-C$ and then substitute in the equation: $$(\alpha x+\beta y)^2=-2fy-2gx-d$$ Let $\exists\lambda\in\mathbb R$ such that : $$(\alpha x+\beta y+\lambda)^2=2(\alpha\lambda-g)x+2(\beta\lambda-f)y+\lambda^2-d$$ Note that the equation of any parabola can be written in the form: $$(distance\;from\;axis)^2=(length\;of\;latus\;rectum)\times(distance\;from\;tangent\;at\;vertex)$$ For the lines on the LHS and RHS to be perpendicular to each other, $$\frac \alpha\beta\times\frac{\alpha\lambda-g}{\beta\lambda-f}=-1\implies \lambda=\frac{\alpha g+\beta f}{\alpha^2+\beta^2}$$ Now we rewrite our equation as: $$\left( \frac{\alpha x+\beta y+\lambda}{\sqrt{\alpha^2+\beta^2}} \right)^2=\frac{2\sqrt{(\alpha\lambda-g)^2+(\beta\lambda-f)^2}}{\alpha^2+\beta^2}\times\frac{2(\alpha\lambda-g)x+2(\beta\lambda-f)y+\lambda^2-d}{2\sqrt{(\alpha\lambda-g)^2+(\beta\lambda-f)^2}}$$
From here we can make several conclusions:
The equation of the axis is:$$\alpha x+\beta y +\lambda=0\tag{i}$$
The equation of tangent at vertex is:$$2(\alpha\lambda-g)x+2(\beta\lambda-f)y+\lambda^2-d=0\tag{ii}$$ From the intersection of above 2 lines, the vertex coordinates are:$$\left( \frac{2f\lambda-\beta\lambda^2-\beta d}{2(\beta g- \alpha f)}\;,\;\frac{2g\lambda-\alpha\lambda^2-\alpha d}{-2(\beta g- \alpha f)} \right)\tag{iii}$$ The lentgh of latus rectum is $$L=\frac{2\sqrt{(\alpha\lambda-g)^2+(\beta\lambda-f)^2}}{\alpha^2+\beta^2}\tag{iv}$$
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, $$\text{(the length of the latus rectum )}=2\times EC=2e=4p=4\times OC.$$
Best Answer
$$y = x \tan \theta - \dfrac{gx^2}{2u^2 \cos^2 \theta }$$
Let $H = \dfrac{u^2 cos^2 \theta}{2g}$ and $R = \dfrac{u^2 \sin 2 \theta}{g}$
So, the given equation transforms into (by multiplying both sides with $\dfrac{2u^2 cos^2 \theta}{g}$ )
$$ 4 Hy = Rx - x^2 $$
$$\left( x - \dfrac{R}{2} \right)^2 = -4H \left(y-\dfrac{R^2}{16H}\right)$$
So, it is actually the parabola in the standard form. So, the length of latus rectum is equal to $4H$ and the vertex is $\left(\dfrac{R}{2}, \dfrac{R^2}{16H} \right)$.
Simplifying those we get,
Vertex : $\left( \dfrac{u^2 \sin 2 \theta }{2g} , \dfrac{u^2 \sin^2 \theta}{2g} \right) $
Length of Latus Rectum : $\dfrac{2u^2 cos^2 \theta}{g}$
BTW: Here $H$ is not eual to the height of projectile.