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?
Let equation of the tangents from point (-3,5) be
(i) posses through (-3,5) so,
Line (i) will be tangent to ellipse `x^2/9+y^2/4=1` if
from (ii) and (iii) ,we have
let angle between lines of (iv) be `theta`
lines (iv) are parallel.
Thus distance between them= `2xx2.9`
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
(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)
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 .