Question

please see question in the attached file

Answer 1

getting there

Answer 2

i get
\[\sqrt{3}\]

Answer 3

the e = c/a ... for an allipse ; is it the same for a hyperbola tho?

Answer 4

@amistre i think (hope) it is
\[\sqrt{1+\frac{b^2}{a^2}}\]

Answer 5

@007 you want to see the work yes?

Answer 6

and you can check it for mistakes

Answer 7

for hyp we get:

\[e=\frac{\sqrt{a^2+b^2}}{a}\]
where a is the transverse vertex and b is the ghost of a departed ellipse :)

Answer 8

yes

Answer 9

ok start by taking the derivative of
\[\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\] wrt x and get
\[\frac{2x}{a^2}-\frac{2yy'}{b^2}=0\] therefore after some algebra i get
\[y'=\frac{b^2x}{a^2y}\]

Answer 10

which is also: \(\cfrac{\text{focal length}}{\text{transverse axis}}\)

Answer 11

and so by substitution at the point (6,3) the slope is
\[\frac{6b^2}{3a^2}=\frac{2b^2}{a^2}\]
which means the normal line has slope
\[-\frac{a^2}{2b^2}\]

Answer 12

P = (6, 3)
xint = -(9, 0)
-----
-3, 3 ; slope = -1 for the normal

Answer 13

and the equation for the normal line is
\[y-3=-\frac{a^2}{2b^2}(x-6)\]
we know it passes through (9,0) so we know
\[0-3=-\frac{a^2}{2b^2}(9-6)\] i.e.
\[-3=-\frac{a^2}{2b^2}\times 3\] and algebra tells us that therefore
\[\frac{a^2}{2b^2}=1\] so
\[\frac{a^2}{b^2}=2\]

Answer 14

and so
\[e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+2}=\sqrt{3}\]

Answer 15

at last i got an answer please check if its correct
differentiating implicitly i got slope of normal as -ya^2/xb^2
and at given point the slope would be -a^2/2b^2
then equation of normal would be \[y-3=-a ^{2}/2b ^{2}(x-6)\]
but this equation satisfies the point (9,0)
then substituting x and y as 9 and 0 in the above equation i got
2b^=a^2
this give \[e =\sqrt{2b ^{2}+b ^{2}/2b ^{2}}\]
and (3/2)1/2

Answer 16

@amistre did you do this in two lines! dang!

Answer 17

@vikki i made a mistake

Answer 18

f(x,y) = (x/a)^2 - (y/b)^2 - 1

fx = 2x/a
fy = 2y/b

the gradient = (2x/a , 2y/b) ; point = (6,3)

(12/a, 6/b) = (-3 , 3) is what ive developed so far, right or wrong :)

Answer 19

a = -4 ; b=2 right?

Answer 20

@amistre i wasn't using partials, just implicit diff but i made a mistake computing e

Answer 21

vikki has it

Answer 22

e = sqrt(4^2+2^2)/4
= sqrt(16 + 4)/4
= sqrt(4(4+1))/4
= 2 sqrt(5)/4
= sqrt(5)/2

if i did it right that is ... which is a big if :)

Answer 23

wow not sure. i think vikki may have it.

Answer 24

it hard to tell, since this window like to auto adjust straight down to the bottom as you read :)

Answer 25

yeah what's up with that?

Answer 26

vikki you have it for sure
it is
\[\frac{b^2}{a^2}=\frac{1}{2}\] so
\[e=\sqrt{1+\frac{1}{2}}=\sqrt{\frac{3}{2}}\]

Answer 27

if i were to adjust my thinking any it would be in scaling the gradient and the normal, if need be ...

Answer 28

you know way more about this that i do. i think i confused the issue. especially since i used a^2/b^2 instead of the other way around

Answer 29

:) if the answer is sqrt(3/2) than I have to find where I missed it at; wolfram says me "e" is right with what I have. But it doesnt match the choices so I have to wonder where I need to adjust it at

Answer 30

f(x,y) = (x/a)^2 - (y/b)^2 - 1

fx = 2x/a^2
fy = -2y/b^2

the gradient = (2x/a^2 , -2y/b^2) ; point = (6,3)

(12/a^2, -6/b^2) = ( 3 , -3)
..............................................

a^2 = 4
b^2 = 2
..........................................

e = sqrt(a^2 + b^2)/a
= sqrt(4+2)/sqrt(4)
= sqrt(6/4)

= sqrt(3/2)

i had forgotten to do my partials correctly :)

Answer 31

IF u happen to know it 8it can be derived algebraically per above) the equation for a normal to hyperbola can be written

x-x1 / x1/a^2 = y-y1 / y1/b^2

so in this case if u set y = 0 and solve for x using (6,3) gives
x = 6 + 6(b^2/a^2) = 9 -> b^2/a^2 = 1/2