Question

If any tangent to the ellipse \frac{X^2}{a^2} + \frac{Y^2}{b^2}=1 \) intercepts equal length \(L\) on the axes then find \( L\)

Sadly it took me more than 5 mints to solve this, how much do you take? ... Am I having intellectual atrophy?

Answer 1

you just have to write the equation if the form Y=F(x)
then find its derivative, equate the derivative equation to -1 and find your X which is your L

Answer 2

what does "intecepts equal length" mean?

Answer 3

intercepts of the tangent are equal, i guess.

Answer 4

It is length of the intercept

Answer 5

so essentially, (0,L) and (L,0) ??

Answer 6

Yes exactly

Answer 7

\[ \frac{X^2}{a^2} + \frac{Y^2}{b^2}=1 \]
\[ \frac{2X}{a^2} + \frac{2Y}{b^2}Y'=0 \]
\[ \frac{X}{a^2} + \frac{Y}{b^2}Y'=0 \]
\[ \frac{Y}{b^2}Y'=-\frac{X}{a^2} \]
\[ Y'=-\frac{b^2}{a^2}\frac{X}{Y} \]

a slope of \(\pm\)45 degrees should create the desired results right?

Answer 8

or when Y' = 1 or -1

Answer 9

(x,y) = (a^2,b^2) or some signed version of it

Answer 10

The answer is sqrt(a^2+b^2) but I dont know how to do it

Answer 11

ah, so not just one point, but a locus of points ... is what that looks like to me

Answer 12

create the equation of the tangent line\[y = -\frac{Xb^2}{Ya^2}(x-X)+Y\]
let (x,y) -> (a,b) be any point on the ellipse
\[b = -\frac{Xb^2}{Ya^2}(a-X)+Y\]
and solve for Y

Answer 13

if memory serves

Answer 14

I did it but in vain

Answer 15

\[b = -\frac{Xb^2}{Ya^2}(a-X)+Y\]
\[b +\frac{Xb^2}{Ya^2}(a-X)=Y\]
\[Ya^2b +Xb^2a-X^2b^2=Y^2a^2\]
\[Xb^2a-X^2b^2=Y^2a^2-Ya^2b\]
i should prolly check to see if ive remembered the gerenal tangent line setup correctly :)

Answer 16

but as is, its starting to resemble a circle equation that you would need to complete the squares on

Answer 17

the general setup for a tangent line from a derivative is: y = f'(a)(x-a) + f(a); for all points (a,f(a)) on f(x)

our intercepts can then be defined as:

when x=0, y = f'(a)a + f(a)
and when y = 0, x = a - f(a)/f'(a)

and you want them to be equal ....

Answer 18

dropped a negative on that y intercept :/ -f'(a)a + f(a)

Answer 19

-f'(a)a + f(a) = a - f(a)/f'(a)

-f'^2(a)a + f'(a)f(a) = f'(a)a - f(a)

a f'^2(a) - (f(a)-a) f'(a) - f(a) = 0

interesting, that turns out to be rather quadratic in general

Answer 20

I got the answer