hard conics question on my test.. (1 Viewer)

[underthesun]

Hyperbola H has equation

x^2/a^2 - y^2/b^2 = 1

eccentricity e

Ellipse E has equation

x^2/(a^2 + b^2) + y^2/b^2 = 1

questions::

a) Show that E has eccentricity 1/e

b) Show that E passes through one focus of H, and H passes through on focus of E

c) draw (bleh)

d) If H and E intersect at P in the first quadrant, show that the acute angle (alpha) between the tangents to the curves at P satisfies this:

tan @ = sqrt(2) * (e + 1/e)

now, @ = alpha

This question appeared in my exam, and i had problems with D.

(btw this question was given to our class as a revision question, but since teacher isn't checking it, no-one bothered to do it. If only.. if only i posted it here earlier..)

dang!

 

[ND]

x^2/a^2 - 1 = y^2/b^2 [from H]
1 - x^2/(a^2+b^2) = y^2/b^2 [from E]
.'. x^2/a^2 = 2 - x^2/(a^2+b^2)
x^2[(2a^2+b^2)/(a^2(a^2+b^2))] = 2
x^2 = 2a^2(a^2+b^2)/(2a^2+b^2)
subbing into E gives:
y^2 = b^4/(2a^2+b^2)
so P[2a^2(a^2+b^2)/(2a^2+b^2), b^4/(2a^2+b^2)]
x_P/y_P = [sqrt2*a*sqrt(a^2+b^2)]/b^2

dy/dx of H = xb^2/ya^2
.'. M_H at P = [sqrt2*a*sqrt(a^2+b^2)]/b^2 * b^2/a^2
= sqrt2*sqrt(a^2+b^2)/a
= sqrt2*e [from part (a)]

similarly M_E at P = -(sqrt2)/e

for an acute angle @:
tan@ = |(M_H - M_E)/(1 + M_H*M_E)|
= |(sqrt2*e + sqrt2/e)/(1 - 2)|
= |-sqrt2*(e + 1/e)|
= sqrt2*(e + 1/e)

 

[underthesun]

that seems like a very quick solution..

Seems that i have a phobia of long-winded questions and end up avoiding them..
on the bright side, no-one in the class did that question, and seems like I'd scab some marks for getting that x^2 value

thanks for solving btw