Question

Find the angle of intersection of the curves x^2+4y^2=32 and x^2-y^2=12 at the point of their intersection

Answer 1

Answer:

(4 , 2) , (4 , -2) , (-4 , 2) , (-4 , -2) .

OR (± 4 , ± 2)

Solution:

Here,

The given equations are :

x² + 4y² = 32 ------(1)

x² - y² = 12 -----(2)

Now,

Subtracting eq-(2) from eq-(1) ,

We have ;

=> (x² + 4y²) - (x² - y²) = 32 - 12

=> x² + 4y² - x² + y² = 20

=> 5y² = 20

=> y² = 20/5

=> y² = 4

=> y = √4

=> y = ± 2

Now,

If y = 2 , then by using eq-(2) ,

We have ;

=> x² - y² = 12

=> x² - 2² = 12

=> x² - 4 = 12

=> x² = 12 + 4

=> x² = 16

=> x = √16

=> x = ±4

Also,

If y = -2 , then by using eq-(2) ,

We have ;

=> x² - y² = 12

=> x² - (-2)² = 12

=> x² - 4 = 12

=> x² = 12 + 4

=> x² = 16

=> x = √16

=> x = ±4

Hence,

Both the curves will intersect at four points , namely ;

(4 , 2) , (4 , -2) , (-4 , 2) , (-4 , -2) .

Or simply (± 4 , ± 2)