Conics Q (1 Viewer)

[CrashOveride]

Prove that the line lx + my + n = 0 touches the ellipse of standard form if a²l² + b²m² = n²

 

[stupid idiot]

Originally posted by CrashOveride
Prove that the line lx + my + n = 0 touches the ellipse of standard form if a²l² + b²m² = n²

(l/-n)x + (m/-n)y = 1

equation of a tangent: xx₁/a² + yy₁/b² = 1

therefore if the line touches, it is at least a tangent to the ellipse, (l/-n) = x₁/a², and (m/-n) = y₁/b².

now substitute l and m into this thing a²l² + b²m² = n²

you will get this thing: x₁²/a² + y²₁/b² = 1

we conclude by saying that since the equation of the tangent to the standard ellipse is xx₁/a² + yy₁/b² = 1 at (x₁, y₁), therefore the line lx + my + n = 0 must touch the standard ellipse since we obtained this result x₁²/a² + y²₁/b² = 1 which shows that equation at (x₁, y₁) satisfy the standard tangent equation... some shit like that.

 

[Crikket]

you solved that, and your name is "stupid idiot"?!?!
very ironic!

 

[ToO LaZy ^*]

ok..i've just had my 2nd lesson in conics and i don't really have a clue about this 'parametric coordinates'...so could someone explain this question to me. (its the 1st of the exercise, so it won't take long )

a.) show that P(acos@, bsin@) lies on the ellipse x²/a + y²/b = 1. (done)
b.) also, if S, S' are the foci, prove that SP = a(1 - ecos@), S'P = a(1 + ecos@)

 

[CrashOveride]

Originally posted by stupid idiot
(l/-n)x + (m/-n)y = 1

equation of a tangent: xx₁/a² + yy₁/b² = 1

therefore if the line touches, it is at least a tangent to the ellipse, (l/-n) = x₁/a², and (m/-n) = y₁/b².

now substitute l and m into this thing a²l² + b²m² = n²

you will get this thing: x₁²/a² + y²₁/b² = 1

we conclude by saying that since the equation of the tangent to the standard ellipse is xx₁/a² + yy₁/b² = 1 at (x₁, y₁), therefore the line lx + my + n = 0 must touch the standard ellipse since we obtained this result x₁²/a² + y²₁/b² = 1 which shows that equation at (x₁, y₁) satisfy the standard tangent equation... some shit like that.

Sorry i ddint understand the last part....
are we like saying coz x₁²/a² + y²₁/b² = 1 ..which shows that at x1, y1 it will satisfy the form of an ellipse,...and since x1/a² = -l/n ...and same typoe thing for -m/n...it also satisfies the equation of that line and thus at x1,y1 its satisfying both thus it touches at x1,y1 ?

If you feel like elaborating...plz do so

 

[sammeh]

the last paragraph of master idiot's answer is just saying that because from manipulation of the eqn lx + my + n = 0 we derived the standard for for the eqn of an ellipse (ie it satisfies the form of an ellipse), it must touch the ellipse. its just a justification - in an exam its wise to explain what you've done if you think it was a little bit left field

 

[ToO LaZy ^*]

Originally posted by ToO LaZy ^*
ok..i've just had my 2nd lesson in conics and i don't really have a clue about this 'parametric coordinates'...so could someone explain this question to me. (its the 1st of the exercise, so it won't take long )

a.) show that P(acos@, bsin@) lies on the ellipse x²/a + y²/b = 1. (done)
b.) also, if S, S' are the foci, prove that SP = a(1 - ecos@), S'P = a(1 + ecos@)

 

[sammeh]

lazy: using the identity that hte foci of an ellipse are (+/- ae, 0) if the ellipses major axis lies on the x axis, ie b^2 = a^2(1 - e^2), just plug into the distance formula and u should get the answer. im too tired at the moment to do it for you, and given said tiredness im probly wrong, but that'd be my first port of call. admittedly, going off on a totally unrelated tangent to the question is a failing of mine ^^

 

[ToO LaZy ^*]

...?

 

[CM_Tutor]

Originally posted by ToO LaZy ^*
a.) show that P(acos@, bsin@) lies on the ellipse x²/a + y²/b = 1. (done)
b.) also, if S, S' are the foci, prove that SP = a(1 - ecos@), S'P = a(1 + ecos@)

For (b) - as with most conics problems - there is the long way and the short way:

Description of Long Way: You know the location of P and S, so find PS², and then note that
distance PS = |(PS²)^1/2|

Short Way: Draw a line through P parallel to the x-axis, meeting the directrix x = a / e at M. By the definition of an ellipse, PS = ePM.

Now, PM = (a / e) - acos @, as PM is horizontal (draw a diagram if you're not convinced).

So, PS = ePM = e[(a / e) - acos @] = a(1 - ecos @), as required.

PM produced meets the other directrix at M', with PS' = ePM'

So, PS' = e[acos @ - (-a / e)] = a(1 + ecos @)

 

[CrashOveride]

How does one prove the area of an ellipse is Pi.a.b. sq. units?

 

[CrashOveride]

Wait, its been resolved

 

[CrashOveride]

The tangent at P(x₁, y₁) on the hyperbola x^2/a^2 - y^2/b^2 = 1 , x₁ > 0, intersects the directrix at Q. S is the focus (ae, 0). Prove PSQ is a right angle.

Any takers ?

 

[CM_Tutor]

Tangent at P(x₁, y₁) to the hyperbola is (xx₁ / a²) - (yy₁ / b²) = 1.

This meets the directrix x = a / e at (a / e, b²(x₁ - ae) / aey₁)

m_PS = y₁ / (x₁ - ae)

m_QS = [b²(x₁ - ae) / aey₁] / [(a / e) - ae]
= [b²(x₁ - ae) / aey₁] * e / a(1 - e²)
= b²(x₁ - ae) / a²y₁(1 - e²)

Now, = m_PS * m_QS = [y₁ / (x₁ - ae)] * [b²(x₁ - ae) / a²y₁(1 - e²)]
= b² / a²(1 - e²)
= -1, as b² = a²(e² - 1), for a hyperbola.

So, PS _|_ SQ, and so angle PSQ = 90°, as required.

 

[CrashOveride]

How did u get the ordinate where it meets the directrix initially? i must have stuffed something up argh its late

 

[CM_Tutor]

The directrix is x = a / e, and hence the x coordinate of Q must be x = a / e.

 

[CrashOveride]

Nah im fine with that
im talking about the ordinate i.e y value (as they call it right?)

 

[CrashOveride]

ahhhh sorry to disturb ya bud
i left out the 1 while i was doing it...i get ya

 

[CrashOveride]

also for a question like this:

Tangent at P(x1,y1) to ellipse 9x^2 + 16y^2 = 144 meets the directricies at T and T' respectively. S and S' are foci, prove that PST and PS'T' are both right angles

I usually do these things the long way....i was thinkign just find the co-ords of T' and T and fine all the neccesary gradients and then do the m1.m2 = -1 thingo and thats how i can show it. Any quick short cuts?

 

[CM_Tutor]

Sorry, I described the wrong one.

Put x = a / e into the equation of the tangent, (xx₁ / a²) - (yy₁ / b²) = 1.

(a / e) * (x₁ / a²) - (yy₁ / b²) = 1
(x₁ / ae) - (yy₁ / b²) = 1
(x₁ / ae) - 1 = yy₁ / b²
(b² / y₁) * [(x₁ / ae) - 1] = y
y = (b² / y₁) * (x₁ - ae) / ae
y = b²(x₁ - ae) / aey₁

So, Q is at (a / e, b²(x₁ - ae) / aey₁), as I stated above.