How do I determine the equation of the tangent to the curve [math] y=2x-x^2 [/math] that passes through point [math] (2,9) [/math]?
There are actually two such tangents:Let y = mx + n be the tangent we're looking for, i.e. we have to determine m and n.By definition of tangent, the line y = mx + n is a tangent to the given curve if they interesect in exactly one point. So let's equate the two equations to calculate intersection(s):[math]y = mx + n = 2x - x^2,[/math][math]x^2 + (m-2)x + n = 0,[/math][math]x = \frac{m-2}{2}\pm\sqrt{\left(\frac{m-2}{2}\right)^2-n}.[/math]If the radicand (the "thing" under the square root) is positive, there will be two intersections. If it is negative, there will be no intersection. If it is zero, there will be one intersection, and the line will indeed be a tangent.So we get:[math]\left(\frac{m-2}{2}\right)^2-n=0,[/math][math]n=\left(\frac{m-2}{2}\right)^2.[/math]Now we can use the fact that the tangent passes through (2,9):[math]9=2m+n=2m+\left(\frac{m-2}{2}\right)^2=2m+\frac{\left(m-2\right)^2}{4},[/math][math]36=8m+4-4m+m^2,[/math][math]m^2+4m-32=0,[/math][math]m=-2\pm\sqrt{4+32}=-2\pm\sqrt{36}=-2\pm 6,[/math][math]m=4\ \text{or}\ m=-8.[/math]For m = 4, we get[math]n=\frac{(2-4)^2}{4}=\frac{4}{4}=1,[/math]and for m = -8, we get[math]n=\frac{(2-(-8))^2}{4}=\frac{100}{4}=25.[/math]The two solutions are therefore y = 4x + 1 and y = -8x + 25.
Find the locus of a point P such that the sum of its distance from A(0,2) and B (0,-2) is 6?
You need to know a distance formula to find the locus.By distance formula,[x^2 + (y-2)^2]^1/2 + [x^2 + (y+2)^2]^1/2 = 6.(x^2 + y^2 - 4y + 4)^1/2 + (x^2 + y^2 + 4y + 4)^1/2 = 6.Let x^2 + y^2 + 4 = a.(a - 4y)^1/2 + (a + 4y)^1/2 = 6.(a - 4y)^1/2 = 6 - (a + 4y)^1/2.Squaring both sides,a - 4y = a + 4y + 36 - 12* (a + 4y)^1/2.8y + 36 = 12* (a + 4y)^1/2.2y + 9 = 3* (a + 4y)^1/2.Squaring again,4y^2 + 36y + 81 = 9a + 36y.9a - 4y^2 - 81 = 0.Putting back the original value of a,9x^2 + 5y^2 - 45 = 0.And we're done!!!