Question

SPORT A vector in the direction of the line tangent to curve of intersection of the paraboloids z = 3x2 + y2 and z = x2 - y2 + 4 at the point (1,1,4) on the curve is given by (1 , a , b). Find a + b.

ASAP

Answer 1

Page 1 First we find intersection points of given surfaces. Given surfaces are 2=3² ty &z=x²-y² + 4 Put z=3x² ty in 22% mit 4, we get, 3x²4 y = x²_ğ t4 ..2x² + 2y3 24 2=3x² + y =3x² + (2-x) = 2x +2=2(6²4) Let = we cost, y=&e sint, ostsen, then 2=2 (2 cost tl) = 4 cost t2 - The parametar representation of the curve C of intersection of given surfaces x=re cost, yo sint, z= 4 cost t2.

Pager which passes through the point (1,1,4) corresponding to the value of pargometer t= 1 / 2 Now that = -Jī sint, hogy a fi cost, Ich denke visine, hit Ti est, et -4 sinit The parametrization of tangent line dt dt a = (-ri sint, de cost,-4 sinzt) which must pass through (1,1,4) corresponding to t = 7 . do =1,1,-4) at 12-1 .: The parametric equations of the vector in the direction of this tangent line are

Pages x= 1+(1) + = 1 + (1)t , z =4 +(-4) lie. x=1-t, y=1tt, an 4-4t. But given that (1-t, 1tt, 4-4+) = (1,4,5) ..l-=1,1+ta, 4-4t=b itão , a=), b=4 ja +b=5]