Solution: The focus of parabola is $(4,0)$. Let slope of focal chord be $m$. Equation of focal chord is $y=m(x-4)$. It is tangent to the circle then

$ \begin{aligned} & \left|\frac{6 m-4 m}{\sqrt{m^{2}+1}}\right|=\sqrt{2} \\ & 4 m^{2}=2\left(m^{2}+1\right) \\ & 2 m^{2}=2 \\ & m= \pm 1 \end{aligned} $

Answer: a

Solution: $\mathrm{ax}=1+\mathrm{by}-2 \sqrt{\mathrm{by}}$

$ \begin{aligned} & (a x-b y-1)^{2}=(-2 \sqrt{b y})^{2} \\ & a^{2} x^{2}+b^{2} y^{2}+1-2 a b x y-2 a x+2 b y=4 b y \\ & a^{2} x^{2}-2 a b x y+b^{2} y^{2}-2 a x-2 b y+1=0 \\ & \begin{array}{rlr} \Delta=\left|\begin{array}{ccc} a^{2} & -a b & -a \\ -a b & b^{2} & -b \\ -a & -b & 1 \end{array}\right| \\ & =a^{2}\left(b^{2}-b^{2}\right)+a b(-a b-a b)-a\left(a b^{2}+a b^{2}\right) \\ & =0-2 a^{2} b^{2}-2 a^{2} b^{2} \\ & =-4 a^{2} b^{2} \neq 0 \end{array} \\ & h^{2}-a b=(-a b)^{2}-\left(a^{2}\right)\left(b^{2}\right)=0 \\ & \therefore \text { It is a parabola } \end{aligned} $

Answer: b

Solution: $\mathrm{y}^{2}+4 \mathrm{y}=-4 \mathrm{x}-2$

$(y+2)^{2}=-4 x-\frac{1}{2}$

$y^{2}=-4 A X$

Equation of directrix is $\mathrm{X}=$ Ai.e. $\mathrm{x}-\frac{1}{2}=1$

$2 \mathrm{x}-3=0$ or $\mathrm{x}=\frac{3}{2}$

Answer: d

Solution: Let $\mathrm{P}\left(\mathrm{at}^{2}, 2 \mathrm{at}\right)$ lies on the parabola

$\mathrm{y}^{2}=4 \mathrm{ax}$

Mid point of PS is Q.

$\mathrm{h}=\frac{\mathrm{at}{ }^{2}+\mathrm{a}}{2}, \mathrm{k}=\frac{0+2 \mathrm{at}}{2}$

$\frac{2 \mathrm{~h}-\mathrm{a}}{\mathrm{a}}=\mathrm{t}^{2}, \frac{\mathrm{k}}{\mathrm{a}}=\mathrm{t}$

$\frac{2 \mathrm{~h}-\mathrm{a}}{\mathrm{a}}=\frac{\mathrm{k}}{\mathrm{a}}^{2}$

$\mathrm{a}(2 \mathrm{~h}-\mathrm{a})=\mathrm{k}^{2}$

Locus of $(h, k)$ is $ y^{2}=2 a x-\frac{a}{2}$

Equaiton of directrix is $x-\frac{a}{2}=-\frac{a}{2}$

$ x=0 $

Answer: c

Solution: Equation of tangent of the parabola $y^{2}=4 x$ is

$ \mathrm{y}=\mathrm{mx}+\frac{1}{\mathrm{~m}} $

This equation passes through $(1,4)$ i.e.

$4=\mathrm{m}+\frac{1}{\mathrm{~m}}$

$ m^{2}-4 m+1=0 $

$\mathrm{m} _{1} \cdot \mathrm{m} _{2}=1$ and $\mathrm{m} _{1}+\mathrm{m} _{2}=4$

Angle between the two tangents is $\tan \theta=\left|\frac{\mathrm{m} _{1}-\mathrm{m} _{2}}{1+\mathrm{m} _{1} \mathrm{~m} _{2}}\right|$

$\tan \theta=\left|\frac{\sqrt{(4)^{2}-4}}{1+1}\right|=\frac{\sqrt{12}}{2}=\sqrt{3}$

$\therefore \theta=\frac{\pi}{3}$

Answer: c

Solution: Equation of tangent at $(1,7)$ to the curve $y=x^{2}+6$ is

$\frac{\mathrm{y}+7}{2}=\mathrm{x}+6$

$2 \mathrm{x}-\mathrm{y}+5=0$

This line also touches the circle i.e.

Equation of normal of circle passing through

centre $(-8,-6)$.

$\mathrm{x}+2 \mathrm{y}+\lambda=0$

$-8-12+\lambda=0$

$\lambda=20$

$\therefore \mathrm{x}+2 \mathrm{y}+20=0$

$\mathrm{Q}$ is intersection point of (1) and (2)

$\mathrm{x}=-6$,

$y=-7$

$\mathrm{Q}(-6,-7)$

Answer: d

Solution: Let eq $\mathrm{q}^{\mathrm{n}}$ of tangent of parabola be $\mathrm{y}=\mathrm{mx}+\frac{1}{\mathrm{~m}}$ is also a tangent to the circle then $\left|\frac{3 m+\frac{1}{m}}{\sqrt{1+m^{2}}}\right|=2 \sqrt{2}$ $\frac{\left(3 m^{2}+1\right)^{2}}{m^{2}}=8\left(1+m^{2}\right)$

$\mathrm{m}^{4}-2 \mathrm{~m}^{2}+1=0$

$\mathrm{m}^{2}=1 \mathrm{~m}= \pm 1 $ Two common tangents are possible.

Solution: We know that $2 \mathrm{a}=\frac{2 \mathrm{xSAxSB}}{\mathrm{SA}+\mathrm{SB}}$

$2 a=\frac{2 b c}{b+c} \Rightarrow \begin{aligned} a b & =b c-a c \\ a b & =(b-o) c\end{aligned}$

$\mathrm{ab}+\mathrm{ac}=\mathrm{bc}$

$\frac{a b}{b-a}=c$

Answer: d

Solution: Let $y=m x+\frac{2}{m}$ be tangent to $y^{2}=8 x$. Since circle intersects the parabola orthogonally. So this tangent is the normal for the circle. Every normal of the circle passes through its centre. So centre $(1,3)$.

$ \begin{aligned} & 3=m+\frac{2}{m} m^{2}-3 m+2=0 \\ & (m-2)(m-1)=0 \\ & m=1,2 \\ & y=x+2 \text { or } y=2 x+1 \\ \end{aligned} $

Answer: a