TS EAMCET 4 May 2018 Shift 1 Solved Paper

The equation of normal to the parabola y2=x having slope m is given as,
y=mx−2t(

1

4

)m−

1

4

m3
The normal is drawn through the point (c,0). So,
0=mc−2(

1

4

)m−

1

4

m3
m3+2m−4mc=0
m[m2+(2−4c)]=0
Thus,
m=0
Or,
m2+2−4c=0
The remaining Two are perpendicular to each other,
12+2−4c=0
2−4c=−1
4c=3
c=

3

4