Let the curves be C_1 : y=5x^3, and, C_2 : y=5x^2C1:y=5x3,and,C2:y=5x2.
To find their pt. of intersection, we have to solve their eqns.
5x^3=y=5x^2 rArr 5x^2(x-1)=0 rArr x=0, x=15x3=y=5x2⇒5x2(x−1)=0⇒x=0,x=1.
y=5x^2, x=0, x=1 rArr y=0, y=5y=5x2,x=0,x=1⇒y=0,y=5. Thus, we have,
C_1nnC_2={O(0,0), A(1,5)}C1∩C2={O(0,0),A(1,5)}.
To find the slopes of tgts. to the curves C_1, and, C_2C1,and,C2, at the pts. O, and, AO,and,A, we will find dy/dxdydx as it denotes the reqd. slopes.
For C_1, dy/dx=15x^2, &, "for" C_2, dy/dx=10xC1,dydx=15x2,&,forC2,dydx=10x
:. for C_1, and, C_2, [dy/dx]_O=0.
This means that, both tgts. to C_1,&,C_2 have the same slope, namely 0, and, both pass thro. the same pt. O(0,0).
Clearly, (1) the X-axis is the common tgt. to C_1 and C_2 touching at the Origin. (2) The angle btwn. them, at O is 0^@.
"At the pt." A(1,5), dy/dx=15, "for" C_1, and, "for" C_2, dy/dx=10
:. m_1=15, and m_2=10, are the respective slopes of tgts.
:. alpha=the /_(C_1,C_2) rArr tan alpha=|(m_1-m_2)/(1+m_1m_2)|
=|(15-10)/(1+150)|=5/151~~0.0331
:. alpha=arc tan 0.0331=1.8965^@.
Enjoy Maths.!