# the tangents on the ellipse (x^2)/9 + (y^2)/4 = 1 are drawn from the point (-3,5). What is the angle they close?

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Let equation of the tangents from point (-3,5) be

y=mx+c (i)

(i) posses through (-3,5) so,

5=-3m+c

c=3m+5 (ii)

Line (i) will be tangent to ellipse `x^2/9+y^2/4=1` if

`c=+-sqrt(9m^2+4)` (iii)

from (ii) and (iii) ,we have

`(3m+5)^2=9m^2+4`

`m=-7/10`

`c=2.9`

y=`-.7x+-2.9` (iv)

let angle between lines of (iv) be `theta`

`tan(theta)=(-.7-(-.7))/(1+(-.7)(-.7))`

`tan(theta)=0`

`theta=0`

lines (iv) are parallel.

Thus distance between them= `2xx2.9`

`=5.8` unit

First of all we note that the ellipse touches the x-axis at the points: x = - 3 e x = 3

explicit respect to x we get: y =2/3 sqrt ( 9 - x^2)

The slope is: y’= . - 2x .

3 sqrt(9 – x2).

A line tangent to the already know, as the abscissa of the point P is one in which the ellipse touches the x axis, then it will be vertical equation x = -3 then the point of tangency will be the point

Q (-3.0) . We have to find the other point R of tangency.

To do this just put the system of equations, one stored at the point R on the ellipse and another of tangency to the same:

x2 + y2 = 1

9 4

(y – y0) = m ( x – x0)

. .

Thus: [ 2 sqrt(9 – x2) – 5 ] = . -2x . ( x + 3)

3 3 sqrt(9 – x2)

that once developed, it leads to:

29x2 – 24x -189 = 0

of solutions: x = - 3 (as we foretold), and x = 63/29 (about 2,17) which R( 63 ; 40)

29

Therefore, the tangent of the angle a we’r searching for is:

tg a = ( 40 -5) / (63/29 + 3) = 0,25

corresponding to the angle: 14° 2’10s .