Conics q..HELP! (1 Viewer)

[DcM]

I need help wif the following conics questions:

1) The point P (a sec#, b tan#) lies one the hyperbola (x^2/a^2) - (y^2/b^2) = 1. Let d1 and d2 be the distances from P to each of the hyperbola's asymptotes. Show d1.d2 = (a^2.b^2)/(a^2 + b^2)


2) The points A (a, 0), A' (-a, 0) and P (a sin#, b cos#) are on the hyperbola (x^2/a^2) - (y^2/b^2) = 1 where a>0. AP and A'P interesect the asymptote ay = bx at Q and R respectively. Show that QR = ae.

Thanks alot!
DcM

 

[CM_Tutor]

Originally posted by DcM
The point P (a sec#, b tan#) lies one the hyperbola (x^2/a^2) - (y^2/b^2) = 1. Let d1 and d2 be the distances from P to each of the hyperbola's asymptotes. Show d1.d2 = (a^2.b^2)/(a^2 + b^2)

The distances d₁ and d₂ are perpendicular distances, and thus should be found using the perpendicular distance formula.

Take d₁ as the distance from p to y = bx / a (ie. bx - ay = 0).
So, d₁ = |b(asec#) - a(btan#)| / sqrt[(b)² + (-a)²] = ab|sec# - tan#| / sqrt(a² + b²)

Similarly, take d₂ as the distance from p to y = -bx / a (ie. bx + ay = 0).
So, d₂ = |b(asec#) + a(btan#)| / sqrt[(b)² + (a)²] = ab|sec# + tan#| / sqrt(a² + b²)

Thus, d₁ * d₂ = [ab|sec# - tan#| / sqrt(a² + b²)] * [ab|sec# + tan#| / sqrt(a² + b²)]
= (ab)²|(sec# - tan#)(sec# + tan#)| / [sqrt(a² + b²)]²
= a²b²|sec²# - tan²#| / (a² + b²)
= a²b²|1| / (a² + b²)
= a²b² / (a² + b²), as required.

 

[CM_Tutor]

Just bringing this back to the top, as Q2 has not yet been answered.

(And, if anyone is wondering, no I haven't spotted any useful geometric short cut...)

 

[CM_Tutor]

Originally posted by DcM
2) The points A (a, 0), A' (-a, 0) and P (a sin#, b cos#) are on the hyperbola (x^2/a^2) - (y^2/b^2) = 1 where a>0. AP and A'P interesect the asymptote ay = bx at Q and R respectively. Show that QR = ae.

We are talking about a hyperbola, so this question should read that P is (asec@, btan@).

To start, we need AP: m_AP = (btan@ - 0)(asec@ - a) = btan@ / a(sec@ - 1)
And, since A(a, 0) lies on AP, its equation is y = b(x - a)tan@ / a(sec@ - 1). The line meets the asymptote ay = bx when: bx / a = b(x - a)tan@ / a(sec@ - 1)
x[1 - tan@ / (sec@ - 1)] = -atan@ / (sec@ - 1)
x = [-atan@ / (sec@ - 1)] * [(sec@ - 1) / (sec@ - 1 - tan@)] = atan@ / (1 + tan@ - sec@)
So, x = asin@ / (sin@ + cos@ - 1)
Since this intersection (at Q) lies on y = bx / a, we easily find that Q is
(asin@ / (sin@ + cos@ - 1), bsin@ / (sin@ + cos@ - 1))

Next, we need R. The equation of A'P is y = b(x + a)tan@ / a(sec@ + 1). The line meets the asymptote ay = bx when: bx / a = b(x + a)tan@ / a(sec@ + 1)
x[1 - tan@ / (sec@ + 1)] = atan@ / (sec@ + 1)
x = [atan@ / (sec@ + 1)] * [(sec@ + 1) / (sec@ + 1 - tan@)] = atan@ / (sec@ + 1 - tan@)
So, x = asin@ / (1 + cos@ - sin@)
Since this intersection (at R) lies on y = bx / a, we easily find that R is
(asin@ / (1 + cos@ - sin@), bsin@ / (1 + cos@ - sin@))

Now, using the distance formula, QR² = (x_Q - x_R)² + (y_Q - y_R)²
= {[asin@ / (sin@ + cos@ - 1) - asin@ / (1 + cos@ - sin@)]² + [bsin@ / (sin@ + cos@ - 1) - bsin@ / (1 + cos@ - sin@)]²}
= (a² + b²)[f(@)]² _____(*), where f(@) = [sin@ / (sin@ + cos@ - 1)] - [sin@ / (1 + cos@ - sin@)]

Now, f(@) = [sin@ / (sin@ + cos@ - 1)] - [sin@ / (1 + cos@ - sin@)]
= sin@[(1 + cos@ - sin@) - (sin@ + cos@ - 1)] / [(1 + cos@ - sin@) * (sin@ + cos@ - 1)]
= 2sin@(1 - sin@) / [(cos@)² - (1 - sin@)²]
= 2sin@(1 - sin@) / [cos²@ - 1 + 2sin@ - sin²@]
= 2sin@(1 - sin@) / [cos²@ - sin²@ - cos²@ + 2sin@ - sin²@], using sin²@ + cos²@ = 1
= 2sin@(1 - sin@) / (2sin@ - 2sin²@)
= 1

So, using (*), QR² = (a² + b²) * (1)² = a² + b²
= a² + a²(e² - 1), noting that b² = a²(e² - 1) for a hyperbola
= a²e²

So, QR = ae, as QR > 0, as required.

 

[McLake]

^That's WAY too much typing ^

 

[victorling]

CM tutor, calm down

 

[George W. Bush]

Woo! Hattrick of useless posts