Conic Intersection Questions
Diapositivas del conjunto (2)
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- 1. y=mx+2 is a tangent to y^2=5x. Find the value of m.
- 2. y=mx-5 is a tangent to the ellipse x^2/4 + y^2/13 = 1
- 3. y^2=6x is a parabola and y=x+3/2 is a tangent to it. Find the point of contact
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- 1. y^2=(mx+2)^2 = m^2x^2+4mx+4. So as me know y^2= this and also 5x, we can say 5x=m^2x^2+4mx+4. We can rearrange this to get (m^2)x^2 + (4m-5)x + 4 = 0. Now as this is a tangent, we know b^2 - 4ac=0 , so (4m-5)^2 - 4(m^2)(4) cancels to give -40m=25, so 40m=25 and therefore m=5/8
- 2. y^2 = (mx-5)^2 = m^2x^2-10mx+25. Rearrange x^2/4 + y^2/13=1 for y^2 : y^2=13-(13/4)x^2. So 13-(13/4)x^2 = m^2x^2-10mx+25 which we rearrange to get (m^2+13/4)x^2 + (-10m) + 12 = 0 , to which we know is a tangent so we use b^2-4ac=0 to find m = +/-root(3)
- 3. y^2 = (x+3/2)^2 = x^2+3x+9/4. We can say 6x = x^2+3x+9/4 to which we rearrange to get x^2-3x+9/4=0 , which factorises to (x-3/2)(x-3/2) so therefore x=3/2. To find y we can use y^2=6x , so root(6x3/2) = 3 so the point of connection is (3/2,3) .