For
A The equation of tangent is given by,
yy1−2(x+x1)=0 So,
√8y=2(x+2) =2x+4 2√2y=2x+4 −√2y+x+2=0 For
B,
The equation of normal is,
y=3x−2am−am3 Here,
y2=16x a=4 m=1 So,
y=x−8−4 x−y−12=0 For
C,
It is given that
y2=12x so,
4a=12 a=3 For the focal chord,
t1t2=−1 t2=− This implies,
x1,y1)=(3t2,6t) and
(x2,y2)=(,−) Then,
y1y2=6t×(−) =−36 For
D y2=kx−16 y2=k(x−) The comparison of above with
y2=4ax gives,
4a=k a= The directrix is given by,
x−=−a x=− Compare the above equation with
x−3=0 then,
−=3 k2+12k−64=0 k=4,−16(k≠−16) Therefore the correct option is,
(a)-(iv), (b)-(vi), (c)-(i), (d)-(ii)