Hey there! Sign in to join this conversationNew here? Join for free

finding intersection points (and other possible problems) Watch

    • Thread Starter
    Offline

    0
    ReputationRep:
    Find the values of k for which the line y = x + 2 meets the curve y2 + (x + k)2 = 2. ( Answer: 0=/<k</= 4 ). can anyone walk me through the steps, please? TA
    Offline

    14
    ReputationRep:
    Square the first equation and substitute into y^2+(x+k)^2=2 to eliminate y.

    As the two lines meet, we know that the resulting equation in x must have real roots. Use this to form a quadratic in k and solve to get two solutions. These will be the inequalities given in the mark scheme. If you're still stuck, post what you have and I'll try and help.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Damask-)
    Square the first equation and substitute into y^2+(x+k)^2=2 to eliminate y.

    As the two lines meet, we know that the resulting equation in x must have real roots. Use this to form a quadratic in k and solve to get two solutions. These will be the inequalities given in the mark scheme. If you're still stuck, post what you have and I'll try and help.
    ah cheers. yeah still a little stuck-- i end up with (after your instructions):
    2x^2+ k^2 + (4+2k)x+2=0

    how am i meant to solve for k now?
    Offline

    14
    ReputationRep:
    (Original post by zomgleh)
    ah cheers. yeah still a little stuck-- i end up with (after your instructions):
    2x^2+ k^2 + (4+2k)x+2=0

    how am i meant to solve for k now?

    So, you've gort a quadratic in x where a=2, b=(4+2k), c=k^2+2.
    What has to be true of the discriminant for this to have real roots?
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Damask-)
    So, you've gort a quadratic in x where a=2, b=(4+2k), c=k^2+2.
    What has to be true of the discriminant for this to have real roots?
    aah ok got it, cheers.

    can you also help me out with this please--

    find a vector in R2 in the direction of the line 2x+7y+3=0 (consider its slope)

    I got the slope as -2/7x and apparently the answer is (-7, 2). I don't get why though?
    • Thread Starter
    Offline

    0
    ReputationRep:
    also would you happen to know the taylor approx about 0 for this function--
    (1+x)^α(alpha)?
    the answer is=
    1+αx+α(α-1)x^2/2+ α(α-1)(α-2)x^3/3!+...
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    What newspaper do you read/prefer?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.