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    • Thread Starter

    Help me with these questions pls.

    1.) The normal to the curve x^2 + 3xy + y^2 = 11 at the point (2,1) meets the axes at the points P and Q. Given that O is the origin, show that the area of triangle OPQ is \frac{81}{112} square units.

    I've done so far:
    x^2 + 3xy + y^2 = 11
    2x + 3x[latex]\frac{dy}{dx} + 3y + 2y\frac{dy}{dx} = 0[/latex]
    \frac{dy}{dx} = \frac{-7}{8}

    Y-Y_1 = m(X-X_1)
    Y-1 = \frac{8}{7} (X-2)
    7Y-8X= -9

    2.) The curve C has parametric equation x=asect, y=btant. prove that \frac{dy}{dx}=\frac{b}{a}cosect. Find the eq. in the form y=px + q of the tangent to C at the point where t=\frac{\pi}{4}.

    Done so far:
    \frac{dy}{dx} = \frac{bsec^2t}{asect tant}

    Hi maltsheys, lets have a look at these questions

    1.) Draw the line 7y - 8x = -9, it might be better to think about it in the form y = mx + c. You should see that there is a clear triangle between the origin, the point where it crosses the x-axis and where it crosses the y-axis. a little hint, the triangle is in the 4th quadrant. Now you know how to calculate the area of a triangle, 0.5 x base x height.

    2.) Ok really think about the equation you have produced, its much simpler if you simply write the equation out in terms of sin(t) and cos(t) and cancel out.
    • Thread Starter

    done it.
    • Thread Starter

    More help pls.

    1.) The parametric equations of a curve are  X \equiv \mathrm{2cos\theta} + \mathrm{cos2\theta}and Y \equiv \mathrm{2sin\theta} + \mathrm{sin2\theta} . Show that stationary values occur on this curve when \mathrm{cos\theta}\equiv \frac{1}{2}.

    I've done \frac{dy}{dx} = 2cosx + 2cos2x over -2sinx -2sin2x

    2.) A curve is given by parametric equations  x\equiv\t^2 and y\equiv\frac{1}{t}

    a.) find the eq. of the tangent at the point A(9,1/3)
    = I've got \frac{dy}{dx}\equiv\frac{-1}{2t^3}
    54y + x = 27 (correct)

    b.) The tangent at A intersects the curve at point B. Find the value of the parameter t at B.
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Updated: February 8, 2010

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