conic sections, ellipse

Question

A particle is travelling clockwise on the elliptical orbit given by $$\displaystyle \frac{x^2}{100} + \frac{y^2}{25} = 1$$ The particle leaves the orbit at the point $(-8, 3)$ and travels in a straight line tangent to the ellipse. At which point will the particle cross the $y$-axis?

You need to tell us what you have tried to do. What are your thoughts on the problem? Questions will be closed if you don't make an effort. — Fly by Night, Aug 21 '14 at 21:15

lab bhattacharjee · Answer 1 · 2013-05-14T10:37:02.090

2

HINT:

From Article $262$ of this, the equation of the tangent at $P(h,k)$ of $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\text{ is }\frac{x\cdot h}{a^2}+\frac{y\cdot k}{b^2}=1$$

So, the equation of the tangent here will be $$\frac{x\cdot (-8)}{100}+\frac{y\cdot 3}{25}=1$$

Now to cross the $y$ axis, $x=0$

edited May 14 '13 at 10:37

answered May 14 '13 at 10:24

lab bhattacharjee

274,582

ok so the particle crosses y at point zero? – precious May 14 '13 at 10:28
cos if i make x=0, th eanswer i get is 25/3. is that the right thing? this question is so frustrating – precious May 14 '13 at 10:30
Just solve for $y$ to get the intercept $\frac{25}{3}$. – Maazul May 14 '13 at 10:30
ok... im so grateful – precious May 14 '13 at 10:30
but there is more, – precious May 14 '13 at 10:31
@precious, "there is more" means? – lab bhattacharjee May 14 '13 at 10:34
Find the equations for (a) the tangent lines, and (b) the normal lines, to the hyperbola y2/4 - x^2\2= 1 when x = 4: – precious May 14 '13 at 10:34
@precious, please ask different Question – lab bhattacharjee May 14 '13 at 10:37
wow, this is great, but the second question has nothing to do with the first. its on hyperbola while the first is on ellipse. the question says: Find the equations for (a) the tangent lines, and (b) the normal lines, to the hyperbola y^2/4 - x^2/2= 1 when x=4: – precious May 14 '13 at 10:41

score 2 · Answer 2 · answered May 14 '13 at 10:36

2

We have $x^2/100+y^2/25=1$ so if we set $F(x,y)=x^2/100+y^2/25-1=0$ then $$y'=m_{\text{tangant} }=\frac{-F_x}{F_y}=\frac{-x}{4y}$$ and so $$m_{(-8,3)}=\frac{8}{12}$$ and the equation of the tangent line as @lab noted is $$y=\frac{8}{12}(x+8)+3$$

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answered May 14 '13 at 10:36

Mikasa

67,374

score 1 · Answer 3 · answered Aug 21 '14 at 19:53

I like to work with homogeneous coordinates.

The ellipse is represented by

$$C = \begin{bmatrix} \frac{1}{100} & 0 & 0 \\ 0 & \frac{1}{25} & 0 \\ 0 & 0 & -1 \end{bmatrix}$$

the 3×3 matrix such that

$$ \left. \begin{pmatrix} x & y & 1 \end{pmatrix} \begin{bmatrix} \frac{1}{100} & 0 & 0 \\ 0 & \frac{1}{25} & 0 \\ 0 & 0 & -1 \end{bmatrix} \begin{pmatrix} x \\ y \\ 1 \end{pmatrix} =0 \right\} \frac{x^2}{100} + \frac{y^2}{25} -1 =0 $$

The tangent point is $P=(-8,3,1)$ (you verify tangency by $P\cdot C P=0$)

The tangent line is $L=C P = (-\frac{2}{25}, \frac{3}{25}, -1)$ with equation

$$ \left. (x,y,1) (-\frac{2}{25}, \frac{3}{25}, -1)=0 \right\} -2x+3y-25=0 $$

You can work out the rest.

conic sections, ellipse

3 Answers3