Question

point of intersection

1. Find the point of intersections of x-y=-4 and x^2 - 32y = 16

2. Find the point of intersections of y= x-1 and x^2 + y^2 = 13

Answer 1

Answer:

y2=4x and x2=32y intersect at (16,8)

at an angle θ

Now slope of the tangent to y2=4x at (16,8)

is given by m1=(dxdy)(16,8)=(2y4)(16,8)

=41

Again slope of tangent to x2=32y at (16,8)

is given by m2=(dxdy)(16,8)=(322x)(16,8)=1.

We know tanθ=m2+m1m2−m1

=1+411−41=53

∴θ=

hope it helps

Answer 2

Answer:

1. x-y=4

=> x=4+y

x^2-32y=16

=> (4+y)^2-32y=16

=> 16+8y+y^2-32y=16

=> y^2-24y=0

=> y(y-24)=0

=> y=0

=> y=24

x=4+y=4+0=4

x=4+y=4+32=36

point of intersection (0,4) / (32,36)

y=x-1

x^2+y^2=13

=> x^2+(x-1)^2=13

=> x^2+x^2-2x+1=13

=> 2x^2-2x+1-13=0

=> 2x^2-2x-12=0

=> x^2-x-6=0

=> (x-3)(x+2)=0

=> x=3

=> x= -2

point of intersection

((3,2 ),( -2,-3)