Question

Prove that the curves 2y2=x3,y=32x cut each other ar right angle at the origin

Answer 1

Answer:

2y^2 = x^3 ------ (1)  

and y^2 = 32x ------- (2)  

Solving (1) and (2), we get  

2 (32x) = x^2  

So ,  x^3 - 64x = 0  

So ,  x (x^2 - 64) = 0

So ,  x = 0 and x = ± 8

As the point of intersection is in the first quadrant it must be positive and should not be equal 0  

So ,  x = 8  

Putting in y^2 = 32 x , we get  

y^2 = 32(8)  

So,  y^2 = 256  

So ,  y = ±8  

Accepting y = 16, we have the point of intersection is P = (8, 16) which lies in the 1st quadrant  

Now Differentiating (1) and (2) w. r. to x, we get

Slope of tangent PT1 (m1) = 3

Slope of tangent PT2 (m2) = 1

Let θ be the acute angle between curves (1) and (2) at P.