TS EAMCET 4 May 2018 Shift 2 Solved Paper

For $A$ The equation of tangent is given by, $y y_{1}-2(x+x_{1})=0$ So, $\sqrt{8} y=2(x+2)$ $=2x+4$ $2\sqrt{2} y=2x+4$ $-\sqrt{2} y+x+2=0$ For $B$, The equation of normal is, $y=3x-2am-am^{3}$ Here, $y^{2}=16x$ $a=4$ $m=1$ So, $y=x-8-4$ $x-y-12=0$ For $C$, It is given that $y^{2}=12x$ so, $4a=12$ $a=3$ For the focal chord, $t_{1} t_{2} = -1$ $t_{2} = -\frac{1}{t_{1}}$ This implies, $(x_{1}, y_{1}) = (3t^{2}, 6t)$ and $(x_{2}, y_{2}) = \left( \frac{3}{t^{2}}, -\frac{6}{t} \right)$ Then, $y_{1} y_{2} = 6t \times \left( -\frac{6}{t} \right)$ $= -36$ For $D$ $y^{2}=k x-16$ $y^{2}=k \left( x - \frac{16}{k} \right)$ The comparison of above with $y^{2}=4 a x$ gives, $4a=k$ $a=\frac{k}{4}$ The directrix is given by, $x-\frac{16}{k}=-a$ $x=\frac{16}{k}-\frac{k}{4}$ Compare the above equation with $x-3=0$ then, $\frac{16}{k}-\frac{k}{4}=3$ $k^{2}+12k-64=0$ $k=4,-16 \;\;\;\; (k \neq -16)$ Therefore the correct option is, (a)-(iv), (b)-(vi), (c)-(i), (d)-(ii)