skillstriker
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Find the equation of the two lines which contain the point (1,3) and are tangents to the parabola y = x2-2x+3
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Quadratic Functions (1 Viewer)
- Thread starter skillstriker
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Carrotsticks
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Let the line be y=mx+b
Substitute into the equation of the parabola and put it in quadratic form yields:
+3-b = 0)
Let the discriminant be equal to 0:
The line passes through (1,3) so we sub that into the equation y=mx+b:
Solve simultaneously with the quadratic we had:
Thus, there exist no real solutions for m.
Therefore, there are no tangents that can be drawn from M which pass through (1,3).
This is also clearly seen via a diagram if you sketch the curve and the point. You will see that the point lies 'inside' the parabola, so no tangents can be drawn.
Substitute into the equation of the parabola and put it in quadratic form yields:
Let the discriminant be equal to 0:
The line passes through (1,3) so we sub that into the equation y=mx+b:
Solve simultaneously with the quadratic we had:
Thus, there exist no real solutions for m.
Therefore, there are no tangents that can be drawn from M which pass through (1,3).
This is also clearly seen via a diagram if you sketch the curve and the point. You will see that the point lies 'inside' the parabola, so no tangents can be drawn.
lol, I think you got the question wrong