Question

find the anglebetween the curve y^2=8x, 4x^2+y^2=32​

Answer 1

Answer:

Step-by-step explanation:

Given curves are

Y²=8x→ equation 1

4x²+y²=32→ equation 2

Substitute equation 1 into equation 2

4x²+8x-32=0

4(x²+2x-8)=0

X²+2x-8=0

X(x+4)-2(x+4)

x +4=0 x-2=0

X=-4 x=4

X =-4

⇒y²=8x

y²=8×-4=-32

y=√-32 ( does not exist)⇄

X=2

⇒y²=8×2=16

y=√16

y =4

Now,

Y²=8x

Diff w.r.to x

2y dy/do =8

dy/dx=4/y

M1= 4/y⇒4/4 (proved above (x,y)is (2,4))

M1 =1

4x²+y²=32

Diff w.r.to x

4(2x) dy/dx= 32 dy/dx

dy/dx=-4(2)/2y

M2=-4(3)/2(2)

=-4(2)/4

=-2

We know that,

Tan theta = 1-( -2)/1+1(-2) (∵m1-m2/1+m1 m2)

= 1+2/-1

=3/-1

Tan theta =3

Theta = Tan -¹ 3

∴proved