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Math Help - Revision Questions

  1. #1
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    Exclamation Revision Questions

    Hi

    I've been doing revision for my yearlies and I've been struggling with these questions:

    1) Prove that, if the difference between the roots of the equation ax^2 + bx + c = 0 is 1, then a^2 = b^2 4ac.

    2) P is a variable on the curve y = 2x^2 + 3 and O is the origin. Q is the point of the section of OP nearer the origin. Find the locus of Q.

    3) Find the equations of the tangents to the curve xy = 6 which are parallel to the line 2y + 3x = 0.

    4) The curves y = 2x^2 x and y^2 = x intersect at (0,0) and at another point Q. Find the angle between the curve at Q.

    Could someone please help me out?

    Thanx a lot!!!

    P.S. I'm really really sorry that I've been posting a lotta questions but I really really need to do good in this exam.
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  2. #2
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    Quote Originally Posted by xwrathbringerx View Post
    Hi

    I've been doing revision for my yearlies and I've been struggling with these questions:

    1) Prove that, if the difference between the roots of the equation ax^2 + bx + c = 0 is 1, then a^2 = b^2 – 4ac.

    2) P is a variable on the curve y = 2x^2 + 3 and O is the origin. Q is the point of the section of OP nearer the origin. Find the locus of Q. Mr F says: Sorry but I can't make sense of this.

    3) Find the equations of the tangents to the curve xy = 6 which are parallel to the line 2y + 3x = 0.

    4) The curves y = 2x^2 – x and y^2 = x intersect at (0,0) and at another point Q. Find the angle between the curve at Q.

    Could someone please help me out?

    Thanx a lot!!!

    P.S. I'm really really sorry that I've been posting a lotta questions but I really really need to do good in this exam.
    1) Use the quadratic forumula to write down the two roots. Take the difference and equate to 1. You get \frac{2 \sqrt{b^2 - 4ac}}{2a} = 1 ....

    --------------------------------------------------------------------------------

    3) 2y + 3x = 0 \Rightarrow y = - \frac{3}{2} x \Rightarrow m = - \frac{3}{2}

    y = \frac{6}{x} \Rightarrow \frac{dy}{dx} = - \frac{6}{x^2}.

    To find the points on the curve whose tangents are parallel to the given line you therefore have to solve the equation \frac{dy}{dx} = - \frac{6}{x^2} = - \frac{3}{2}.

    You should be able to write down the equation of the tangent knowing the gradient of the tangent and a point on the tangent.

    -----------------------------------------------------------------------------------

    4) Find the coordinates of Q by solving y = 2 (y^2)^2 - y^2. Noting that tangents are lines, you need to find the angle between these two tangents by using the formula for the angle between two lines (which means you will first need to get gradient of the tangents to the two curves at Q): http://www.tpub.com/math2/5.htm.
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  3. #3
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    Exclamation

    oops culd the question possibly b referring to the trisection? "Q is the point of trisection of OP nearer the origin."
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