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M(-1, 4) and N(2, -3) are two fixed points. A variable point S(x, y) moves such that 2|MS| = |NS|. Find the equation of the locus of S. Expressed in its simplest form and state its constant properties.

Mathematics

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Answer

(0, 11)

(0,53)\left(0, \frac{5}{3}\right)

Step-by-step solution

4 steps

  1. To find the points where the locus meets the y-axis, set x=0x=0 in the equation of the locus found in part (a).

    The equation of the locus is 3x2+3y2+12x−38y+55=03x^2 + 3y^2 + 12x - 38y + 55 = 0.
    Substitute x=0x=0:

    3(0)2+3y2+12(0)−38y+55=03(0)^2 + 3y^2 + 12(0) - 38y + 55 = 0 3y2−38y+55=03y^2 - 38y + 55 = 0

  2. Solve the quadratic equation for yy using the quadratic formula y=−b±b2−4ac2ay = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.

    Here, a=3a=3, b=−38b=-38, and c=55c=55.

    y=−(−38)±(−38)2−4(3)(55)2(3)y = \frac{-(-38) \pm \sqrt{(-38)^2 - 4(3)(55)}}{2(3)} y=38±1444−6606y = \frac{38 \pm \sqrt{1444 - 660}}{6} y=38±7846y = \frac{38 \pm \sqrt{784}}{6}

  3. Calculate the square root and find the two values for yy.

    784=28\sqrt{784} = 28 y=38±286y = \frac{38 \pm 28}{6}

    The two possible values for yy are:

    y1=38+286=666=11y_1 = \frac{38 + 28}{6} = \frac{66}{6} = 11 y2=38−286=106=53y_2 = \frac{38 - 28}{6} = \frac{10}{6} = \frac{5}{3}

  4. State the points where the locus meets the y-axis.

    Since x=0x=0, the points are (0,y1)(0, y_1) and (0,y2)(0, y_2).
    The points are (0,11)\boxed{(0, 11)} and (0,53)\boxed{\left(0, \frac{5}{3}\right)}.

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