Question

find the angle of intersection of the curve x^2+y^2=8 and x^2-2y^2=4​

Answer 1

Answer:

Given curves are

x

2

+4y

2

=8, ..............(1)

x

2

−2y

2

=2 ...............(2)

2x+8yy

′

=0, 2x−4yy

′

=0

y

′

=−

8y

−2x

y

′

=

4y

2x

(m

1

) (m

2

)

Also, point of intersection,

x

2

=8−4y

2

(from (1))

Put in 2

8−4y

2

−2y

2

=2

6y

2

=6

y

2

=1

⇒y=+1,−1x=4,4

Point (4,1) and (4,−1)

The angle of the intersection of two curves is the angle between tangent at their point of intersection.

Let m

1

and m

2

be the slope of tangent at point of intersection

m

1

=−

8y

2x

=−

4y

x

m

2

=

4y

2x

=

2y

x

for (4,1)

m

1

=

4

−4

=−1

m

2

=

2

4

=2

tanθ=

∣

∣

∣

∣

∣

1+m

1

m

2

m

1

−m

2

∣

∣

∣

∣

∣

=

∣

∣

∣

∣

∣

1−2

−1−2

∣

∣

∣

∣

∣

=3

θ=tan

−1

3