Position of a Point W.R.T Parabola

Q. The equation of a parabola is $y^2=4x$. Let P (1, 3) and Q (1, 1) are two points in the xy plane. Then,

1. P and Q are exterior points
2. P is an interior point while Q is an exterior point
3. P and Q are interior points
4. P is an exterior point while Q is an interior point

Q. Let the curve $C$ be the mirror image of the parabola $y^2=4x$ with respect to the line $x+y+4=0$. If $A$ and $B$ are the points of intersection of $C$ with the line $y=–5$, then the distance between $A$ and $B$ is

Q. Consider the parabola, $(y+3{)}^2=2(x-5)$, then the correct option(s) among the following is/are

1. point $(5,-3)$ lies on the parabola
2. point $(2,3)$ lies inside the parabola
3. point $(0,4)$ lies outside the parabola
4. point $(8,-2)$ lies inside the parabola

Q. The point $(-2m,m+1)$ is an interior point of the smaller region bounded by the circle $x^2+y^2=4$ and the parabola $y^2=4x$. Then $m$ belongs to the intyerval

1. $-5-2\sqrt{6}<m<1$
2. $0<m<4$
3. $-1<m<\frac{3}{5}$
4. $-1<m<-5+2\sqrt{6}$

Q. If $(\alpha +1,\alpha )$ be a point interior to the regions of the parabola $y^2=4x$ bounded by the chord joining the points $(6,5)$ and $(7,4),$ then the total number of integral values of $\alpha$ is

Q. If $({\alpha }^2,\alpha -2)$ be a point interior to the regions of the parabola $y^2=2x$ bounded by the chord joining the points $(2,2)$ and $(8,-4),$ then $\alpha$ belongs to the interval

1. $-2+2\sqrt{2}<\alpha <2$
2. $\alpha >-2+2\sqrt{2}$
3. $\alpha >-2-2\sqrt{2}$
4. $\alpha <-2-2\sqrt{2}$

Q. If point $(k,k^2)$ lies inside the region bounded by parabolas $y^2=64x$ and $-x^2+x-1+y=0$ then $k$ lies in the interval

1. $k\in (1,4)$
2. $k\in (0,4)$
3. $k\in (1,\infty )$
4. $k\in (-\infty ,4)$

Q. If $(\alpha +1,\alpha )$ be a point interior to the regions of the parabola $y^2=4x$ bounded by the chord joining the points $(6,5)$ and $(7,4),$ then the total number of integral values of $\alpha$ is

Q. The point $(-2m,m+1)$ is an interior point of the smaller region bounded by the circle $x^2+y^2=4$ and the parabola $y^2=4x$, then $m$ lies in the interval

1. $-5-2\sqrt{6}<m<1$
2. $0<m<4$
3. $-1<m<\frac{3}{5}$
4. $-1<m<-5+2\sqrt{6}$