$\text{Let}{\mathrm{ A}}_1,B_1,C_1\text{be three points in the}xy-\text{plane.}\text{Suppose that the lines}{A}_1{C}_1\text{and}{B}_1{C}_1\text{are tangents to the}$

$\text{curve}{\mathrm{ y}}^2=8x\text{at}{\mathrm{ A}}_1\text{and}{B}_1,\text{respectively.}\text{If}\mathrm{ O}=(0,0)\text{and}{C}_1=(-4,0),\text{then which of the following}$

$\text{statements is(are) TRUE?}$

Correct Options:

• The length of the line segment O$A_1$ is $4\sqrt{3}$

• The orthocenter of the triangle $A_1{B}_1{C}_1$ is (0, 0)

Step-by-Step Explanation:

Step 1: Parametric Representation of the Parabola

The equation of the given parabola is:

$y^2=8x$

Comparing this with the standard equation of a parabola $y^2=4ax$, we get:

$4a=8\Rightarrow a=2$

Any point on the parabola in parametric form can be written as $(a{t}^2,2at)=(2t^2,4t)$.

Let the points of contact be:

$A_1=(2t_1^2,4t_1)$

$B_1=(2t_2^2,4t_2)$

Step 2: Finding the Coordinates of Points $A_1$ and $B_1$

The intersection point $C_1$ of the tangents at $A_1$ and $B_1$ is given by:

$C_1=(a{t}_1{t}_2,a(t_1+t_2\left)\right)=(2t_1{t}_2,2(t_1+t_2\left)\right)$

We are given that $C_1=(-4,0)$. Equating the corresponding coordinates:

$2t_1{t}_2=-4\Rightarrow t_1{t}_2=-2$

$2(t_1+t_2)=0\Rightarrow t_2=-t_1$

Substituting $t_2=-t_1$ into $t_1{t}_2=-2$:

$t_1(-t_1)=-2\Rightarrow t_1^2=2\Rightarrow t_1=\sqrt{2} \text{and} t_2=-\sqrt{2}$

Now, substitute these parameters back into the parametric points:

$A_1=\left(2\right(\sqrt{2}{)}^2,4\sqrt{2})=(4,4\sqrt{2})$

$B_1=\left(2\right(-\sqrt{2}{)}^2,4(-\sqrt{2}))=(4,-4\sqrt{2})$

Step 3: Checking Option 1 (Length of Line Segment $O{A}_1$)

Using the distance formula between the origin $O(0,0)$ and $A_1(4,4\sqrt{2})$:

$O{A}_1=\sqrt{4^2+(4\sqrt{2}{)}^2}=\sqrt{16+32}=\sqrt{48}=4\sqrt{3}$

Hence, the statement "The length of the line segment $O{A}_1$ is $4\sqrt{3}$" is TRUE.

Step 4: Checking Option 3 (Orthocenter of Triangle $A_1{B}_1{C}_1$)

The vertices of triangle $A_1{B}_1{C}_1$ are $A_1(4,4\sqrt{2})$, $B_1(4,-4\sqrt{2})$, and $C_1(-4,0)$.

1. Since both $A_1$ and $B_1$ have an x-coordinate of $4$, the side $A_1{B}_1$ is a vertical line $x=4$.

Therefore, the altitude from vertex $C_1(-4,0)$ to side $A_1{B}_1$ must be a horizontal line passing through $C_1$, which is the x-axis:

$y=0$

2. Next, let's find the slope of side $B_1{C}_1$:

$\text{Slope of }{B}_1{C}_1=\frac{0-(-4\sqrt{2})}{-4-4}=\frac{4\sqrt{2}}{-8}=-\frac{1}{\sqrt{2}}$

The altitude from $A_1(4,4\sqrt{2})$ to $B_1{C}_1$ is perpendicular to $B_1{C}_1$, so its slope is $\sqrt{2}$.

The equation of this altitude is:

$y-4\sqrt{2}=\sqrt{2}(x-4)\Rightarrow y-4\sqrt{2}=\sqrt{2}x-4\sqrt{2}\Rightarrow y=\sqrt{2}x$

3. Finding the intersection of the two altitudes ($y=0$ and $y=\sqrt{2}x$):

$0=\sqrt{2}x\Rightarrow x=0, y=0$

Thus, the orthocenter of triangle $A_1{B}_1{C}_1$ is $(0,0)$.

Hence, the statement "The orthocenter of the triangle $A_1{B}_1{C}_1$ is (0, 0)" is TRUE.