Equation of parabola whose latusrectum coincide with ellipse.

Question

Finding equation of parabola which latusrectum coincide with the latusrectum of ellipse $\displaystyle \frac{x^2}{25}+\frac{y^2}{16}=1$

Try: Eccentricity $16=25(1-e^2)$

We have $\displaystyle e=\frac{3}{5}$. So focus $(\pm 3,0)$

So equation of latusrectum of ellipse is $x=\pm 3$

Could some help me to solve it, thanks

Answer 1

One method can be the use of general formula for conic sections.General formula for ellipse, hyperbola and parabola is:

$y^2=2px -(1-e^2)x^2$

Where the vertex A of the conic section is supposed to be coincided with origin and their focuses coincided. $e=1$ for parabola and formula reduces to $y^2=2px$. Here Latus rectum of ellipse and parabola are coincided, assuming p for parabola has same value as of ellipse, we can calculate it as follows:

$p=a(1-e^2)$

where e is the eccentricity of ellipse, as you found is $e=\frac{3}{5}$ and $a=5$

⇒ $p=5[1-(3/5)^2]=\frac{16}{5}$

Therefore the equation of parabola must be:

$y^2= 2 \times \frac{16}{5}\times x=\frac{32}{5}x$

This is for when origin is at vertex of parabola. The vertex of ellipse is at a distance like $\delta + a$ where $a=5$ and $\delta $ is the distance between the focus and vertex of ellipse(or parabola):

$\delta=\frac{p}{1+e}=\frac{16/5}{1+3/5}=2$ ⇒ $\delta +a=2+5=7$, therefore the equation of prabola must be:

$y^2 =\frac{32}{5}(x-7)$

Notice: For more details of general formula see "Higher mathematics, M.Vygodsky, Mir publishing, 1975, page 69 - 77.

Answer 2

The latus rectum is not a line, but a line segment, so you have to find the intersection of the lines that you’ve determined with the ellipse. What you really want here is the length of the semi-latus rectum, which will just be the absolute value of the $x$-coordinates of these intersection points. This length is also the distance between the parabola’s focus and directrix. Together with the fact that the latus rectum is parallel to the directrix, you should now be able to construct an equation for the parabola. Since you have two points on the parabola and its focus and axis direction, you could instead use those directly to construct an equation.

Note that there are four possible solutions, which reflect the choice of the ellipse focus through which you draw the latus rectum and the direction in which the parabola opens.