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math is hard

Guest

parabola question

hi, i've done parts (a) and (b) but not (c) and (d). (d) is easy but to do it i need to know the tangent lines from (c).

"The diagram shows the parabola y² = 4x and the point A with co-ordinates (3,4).

(a) Find an equation of the straight line having gradient m and passing through the point A(3,4)."

that should just be y - 4 = m(x - 3)...

(b) Show that, if this straight line intersects the parabola, then the y-coordinates of the points of intersection satisfy the equation

my² - 4y + (16 - 12m) = 0

did this too. just re-arrange first equation to get x = (y - 4 + 3m)/(m), then substitute that into equation y² = 4x to get

my² - 4y + (16 - 12m) = 0. easy enough..

(c) By considering the discriminant of the equation in part (b), find the equations of the two tangents to the parabola which pass through A. perpendicular to the tangent at Q.

now THIS is where i am completely stuck and do not know what to do. i also don't know where "Q" is because on the diagram theres just a picture of the parabola and the point A. no mention of Q anywhere else in the question. also, i dont know how to find the equations of the tangents... i am guessing i need to solve the equation:

my² - 4y + (16 - 12m) = 0

for y. i did this and got

y = ( 2 ± 2√[(3m-1)(m-1)] ) / (m)

but i have no idea wht to do with this. thank for your help.

math is hard

Guest

Re: parabola question

nevermind i got it.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: parabola question

Hi math is hard;

nevermind i got it.

You might want to change that username to math is easy. Ain't that the greatest feeling in the world? When the answer just comes in a flash.

Welcome to the forum!

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