# The focus and length of the latus rectum of the parabola $y = x \tan \theta - \frac{gx^2}{2u^2 \cos^2 \theta}$

$$y = x \tan \theta - \frac{gx^2}{2u^2 \cos^2 \theta}$$

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?

• Hint: The formulae you know can be applied to parabolas of the form $4a(y-c) = (x-d)^2$ by translation. Now try to complete the square on the RHS you've given. – stochasticboy321 Oct 17 '15 at 7:15
• Try shifting the origin. – Aditya Kumar Oct 17 '15 at 16:25

$$y = x \tan \theta - \dfrac{gx^2}{2u^2 \cos^2 \theta }$$

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.