Question 306.
Find the equation of the parabola with the vertical axis that passes through the point (0,2) and the points of intersection of the parabolas x^2+2x+3y+4=0 and x^2-3x+y+3=0.
Answer 306.
Equation of a parabola passing through the points of intersection of the parabolas
x^2 + 2x + 3y + 4 = 0 and x^2 - 3x + y + 3 = 0 is given by
x^2 + 2x + 3y + 4 + k (x^2 - 3x + y + 3) = 0 ... (1)
As it passes through (0, 2),
0 + 0 + 3*2 + 4 + k (0 - 0 + 2 + 3) = 0
=> k = - 2
Plugging the value of k = - 2 in (1),
the equation of the required parabola is
x^2 + 2x + 3y + 4 - 2 (x^2 - 3x + y + 3) = 0
=> x^2 - 8x - y + 2 = 0.
Wolfram Alpha Link
Link to YA!
Find the equation of the parabola with the vertical axis that passes through the point (0,2) and the points of intersection of the parabolas x^2+2x+3y+4=0 and x^2-3x+y+3=0.
Answer 306.
Equation of a parabola passing through the points of intersection of the parabolas
x^2 + 2x + 3y + 4 = 0 and x^2 - 3x + y + 3 = 0 is given by
x^2 + 2x + 3y + 4 + k (x^2 - 3x + y + 3) = 0 ... (1)
As it passes through (0, 2),
0 + 0 + 3*2 + 4 + k (0 - 0 + 2 + 3) = 0
=> k = - 2
Plugging the value of k = - 2 in (1),
the equation of the required parabola is
x^2 + 2x + 3y + 4 - 2 (x^2 - 3x + y + 3) = 0
=> x^2 - 8x - y + 2 = 0.
Wolfram Alpha Link
Link to YA!
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