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2

The chord of circle x^2 + y^2 = 25 connecting A(0,5) and B(4,3) is the base of an equilateral triangle with the third vertex C in the first quadrant. What is the smallest possible x-coordinate of C?

CW
Featured 4 days ago

Smallest possible x-co-ordinate of $C = \left(2 - \sqrt{3}\right)$

Explanation:

solution 2 :

Given a circle ${x}^{2} + {y}^{2} = 25 , \implies \text{ center " O=(0,0), and "radius } r = 5$ units
let $D$ be the midpoint of chord $A B$,
$\implies O D$ is perpendicular to $A B$,
as $A C = B C$ and $D$ is midpoint of $A B , C$ lies on line $O D$
$A B = \sqrt{{4}^{2} + {2}^{2}} = \sqrt{20} = 2 \sqrt{5}$ units
$O D = \sqrt{{2}^{2} + {4}^{2}} = \sqrt{20} = 2 \sqrt{5}$ units
$C D = A C \sin 60 = A B \sin 60 = 2 \sqrt{5} \cdot \frac{\sqrt{3}}{2} = \sqrt{15}$
$\implies O C = O D - C D = \sqrt{20} - \sqrt{15}$ units
slope of $O D = {m}_{O D} = \frac{4}{2} = 2$
$\tan \alpha = 2 , \implies \sin \alpha = \frac{2}{\sqrt{5}} , \cos \alpha = \frac{1}{\sqrt{5}}$
Let co-ordinates of $C = \left(x , y\right)$
$\implies C \left(x , y\right) = \left(O C \cos \alpha , O C \sin \alpha\right)$
$= \left(\begin{matrix}\sqrt{20} - \sqrt{15} \cdot \frac{1}{\sqrt{5}} \\ \text{ } \sqrt{20} - \sqrt{15} \cdot \frac{2}{\sqrt{5}}\end{matrix}\right)$
=(sqrt4-sqrt3, " " 2(sqrt4-sqrt3)
$= \left(2 - \sqrt{3} , \text{ } 4 - 2 \sqrt{3}\right)$

Hence, smallest possible x-co-ordinate of $C = \left(2 - \sqrt{3}\right)$

1

Find the value of n??

Jim G.
Featured 1 week ago

$n = 5$

Explanation:

$\text{using the "color(blue)"Intersecting chords theorem}$

$\text{When 2 chords intersect inside a circle, the product of}$
$\text{their segments are equal}$

$\Rightarrow 18 \left(n + 1\right) = 9 \left(n + 7\right)$

$\text{divide both sides by 9}$

$\Rightarrow 2 \left(n + 1\right) = n + 7$

$\Rightarrow 2 n + 2 = n + 7$

$\text{subtract "n" from both sides}$

$2 n - n + 2 = 7$

$\Rightarrow n + 2 = 7$

$\text{subtract 2 from both sides}$

$\Rightarrow n = 7 - 2 = 5$

1

Circle A has a radius of 2  and a center of (3 ,1 ). Circle B has a radius of 6  and a center of (8 ,5 ). If circle B is translated by <-4 ,-1 >, does it overlap circle A? If not, what is the minimum distance between points on both circles?

F. Javier B.
Featured 1 week ago

See process below

Explanation:

Equation for $A$....${\left(x - 3\right)}^{2} + {\left(y - 1\right)}^{2} = {2}^{2}$

Equation for $B$....${\left(x - 8\right)}^{2} + {\left(y - 5\right)}^{2} = {6}^{2}$

A traslation is a isometric transformation, then the image for B in this movement is other circle of radius 6 and center in $\left(4 , 4\right)$. Thus we have the new equation for B is ${\left(x - 4\right)}^{2} + {\left(y - 4\right)}^{2} = {6}^{2}$

Obviously doesn`t overlap both circles and the minimum distance will lay in the line joining both centers

Lets calculate line equation

$\frac{x - 4}{1} = \frac{y - 4}{3}$ or $3 \left(x - 4\right) = y - 4$ then $y = 3 x - 8$
This line intercept to both circles in 4 points (see graph)

Distance is calculated by the standard formula (euclidean distance)

Lets see these points:
$A = \left(\frac{\sqrt{10}}{5} + 3 , 3 \frac{\sqrt{10}}{5} + 1\right)$
$B = \left(- \frac{\sqrt{10}}{5} + 3 , - 3 \frac{\sqrt{10}}{5} + 1\right)$
$C = \left(3 \frac{\sqrt{10}}{5} + 4 , 9 \frac{\sqrt{10}}{5} + 4\right)$
$D = \left(- 3 \frac{\sqrt{10}}{5} + 4 , - 9 \frac{\sqrt{10}}{5} + 4\right)$

The minimum distance will be $B D = - \sqrt{10} + 4$

2

Circle A has a radius of 2  and a center of (3 ,1 ). Circle B has a radius of 6  and a center of (8 ,5 ). If circle B is translated by <-4 ,-1 >, does it overlap circle A? If not, what is the minimum distance between points on both circles?

Somebody N.
Featured 1 week ago

color(blue)(4-sqrt(10)color(white)(888)"units"

Explanation:

The general equation of a circle is:

${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$

Where:

$\boldsymbol{h}$ and $\boldsymbol{k}$ are the $x$ and $y$ coordinates of the centre respectively, and $\boldsymbol{r}$ is the radius.

Circle A

${\left(x - 3\right)}^{2} + {\left(y - 1\right)}^{2} = 4$

If B is translated by $\left(\begin{matrix}- 4 \\ - 1\end{matrix}\right)$, then its centre is translated by this.

Centre:

$\left(\begin{matrix}8 \\ 5\end{matrix}\right) + \left(\begin{matrix}- 4 \\ - 1\end{matrix}\right) = \left(\begin{matrix}4 \\ 4\end{matrix}\right)$

Circle B:

${\left(x - 4\right)}^{2} + {\left(y - 4\right)}^{2} = 36$

By using the distance between the centres and the radii we can deduce the following:

Let:

$d = \text{distance between centres}$

${r}_{1} , {r}_{2} = \text{radii}$

If:

$d > {r}_{1} + {r}_{2}$ Then the circles do not touch.

$d < {r}_{1} + {r}_{2}$ Then the circles intersect at two points or one circle is contained in the other.

$d = {r}_{1} + {r}_{2}$ Then the circles touch at one point.

Using the distance formula:

$d = \sqrt{{\left(3 - 4\right)}^{2} + {\left(1 - 4\right)}^{2}} = \sqrt{10}$

$2 + 6 = 8$

$\sqrt{10} < 8$

So the circles intersect at two points or one is contained in the other. This can be tested by noticing that if the diameter of the smaller circle is less than the radius of the larger then the smaller circle is contained in the larger one.

Diameter of smaller circle is $4$

Radius of larger circle is $6$

So smaller circle is contained in the larger.

To find the shortest distance:

"radius of B"-("radius of " A+d)

$6 - \left(2 + \sqrt{10}\right) = 4 - \sqrt{10} \textcolor{w h i t e}{888} \text{units}$

PLOT:

2

dk_ch
Featured 5 days ago

option (1) $6 \sqrt{2}$

Explanation:

Q92

Given that G is the centroid of $\Delta A B C$

$G A = 2 \sqrt{3} , G B = 2 \sqrt{2} \mathmr{and} G C = 2$

AP median is produced such that $G P = P O$

$B , O \mathmr{and} C , O$ are joined. The quadrilateral CBCO is a parallelogram,

As $A G : G P = 2 : 1$ we get

$\Delta B G P = \frac{1}{3} \Delta A B P = \frac{1}{3} \cdot \frac{1}{2} \Delta A B C$

Again $\Delta B G O = 2 \Delta B G P$

So $\Delta A B C = 3 \Delta B G O$

Now $G O = G A = 2 \sqrt{3}$

$B O = G C = 2$

and $G B = 2 \sqrt{2}$

So $G {O}^{2} - {\left(2 \sqrt{3}\right)}^{2} = 12$

$B {O}^{2} + B {G}^{2} = {2}^{2} + {\left(2 \sqrt{2}\right)}^{2} = 12$

So $\Delta B G O$ is rigtangled at B

So DeltaBGO=1/2BO×BG
Area of

$\Delta A B C = 3 \Delta B G O = \frac{3}{2} \times B O \times B G = \frac{3}{2} \times 2 \times 2 \sqrt{2} = 6 \sqrt{2}$

1

The chord of circle x^2 + y^2 = 25 connecting A(0,5) and B(4,3) is the base of an equilateral triangle with the third vertex C in the first quadrant. What is the smallest possible x-coordinate of C?

Dean R.
Featured 4 days ago

$C = \left(2 - \sqrt{3} , 4 - 2 \sqrt{3}\right)$

Explanation:

We're oddly given a circle in this problem and awkward wording asking for the smallest possible x coordinate.

I'm not sure what the circle has to do with anything. We could list equations endlessly for circles containing these two points. Forget about the circle.

Given one side, there will be two equilateral triangles with that as a side. Let's rewrite the problem:

Find all possible third vertices $C \left(x , y\right)$ which make an equilateral triangle with $A \left(a , b\right)$ and $B \left(c , d\right)$.

There are a few different ways to do this. I'd lean toward complex numbers, but I probably should try to keep this simpler than that.

Let's review. The altitude $h$ of an equilateral triangle bisects a side $s$ so ${\left(\frac{s}{2}\right)}^{2} + {h}^{2} = {s}^{2}$ or $h = \setminus \frac{\sqrt{3}}{2} s$

We just need to go $\pm h$ along the perpendicular bisector of AB and we'll get to our two possible Cs.

The midpoint D of AB is $D \left(\frac{a + c}{2} , \frac{b + d}{2}\right)$

The direction vector from A to B is $B - A = \left(c - a , d - b\right)$.

For the perpendicular direction vector we swap and negate one:

$P = \left(d - b , a - c\right)$

Now schematically what we're doing is

$C = D \setminus \pm h \frac{P}{|} P | = D \setminus \pm \frac{h}{|} P | \setminus P$

We have $s = | A B | = | P |$ so $\frac{h}{|} P | = \frac{\sqrt{3}}{2}$

$C = D \pm \frac{\sqrt{3}}{2} P$

$C = \left(\frac{a + c}{2} , \frac{b + d}{2}\right) \pm \frac{\sqrt{3}}{2} \left(d - b , a - c\right)$

That's the general solution; let's apply it to

$\left(a , b\right) = \left(0 , 5\right) , \left(c , d\right) = \left(4 , 3\right)$

$C = \left(\frac{0 + 4}{2} , \frac{5 + 3}{2}\right) \pm \frac{\sqrt{3}}{2} \left(- 2 , - 4\right)$

$C = \left(2 , 4\right) \pm \sqrt{3} \left(1 , 2\right)$

$C = \left(2 + \sqrt{3} , 4 + 2 \sqrt{3}\right) \mathmr{and} \left(2 - \sqrt{3} , 4 - 2 \sqrt{3}\right)$

Uh, the one with the least x coordinate is

$C = \left(2 - \sqrt{3} , 4 - 2 \sqrt{3}\right)$

Check:

We check the squared distance to each point

$| C - A {|}^{2} = {\left(2 - \sqrt{3}\right)}^{2} + {\left(4 - 2 \sqrt{3} - 5\right)}^{2} = 7 - 4 \sqrt{3} + 13 + 4 \sqrt{3} = 20$

 |C-B|^2 = (2-sqrt{3}- 4 )^2 + (4-2sqrt{3} - 3 )^2 = 20 quad sqrt

1

When a rectangle is rotated about a line it becomes a rectangular prism. Do you agree? Im not sure. It might be a cylinder.

Somebody N.
Featured 3 days ago

See below.

Explanation:

This question is a little ambiguous. I'm guessing you mean a volume of revolution. If this is the case then a rectangle rotated about a line could be a solid cylinder or a hollow cylinder ( Tube) if the line of rotation is parallel or perpendicular to the sides of the rectangle. If the line of rotation is neither parallel nor perpendicular then a vast number of different solids could result. A big problem here is on the definition. Strictly a prism has a base that is a polygon and faces that are flat, but you could say a cylinder has a polygonal base with an in finite number of sides and an infinite number of faces.

Here are two rotations of a rectangle around different lines.

$y = 0$

$x = - 2$

1

A triangle has vertices at A(a,b ), C(c,d), and O(0,0). What are the endpoints and length of the perpendicular bisector of AC ?

Dean R.
Featured 2 days ago

Hey, this got featured before it was really done. It's still not done.

Explanation:

This is another in my series where I generalize one of these old questions.

"The triangle's perpendicular bisectors" is not a term we hear that often. Each side has a perpendicular bisector, the perpendicular through the midpoint. It will occasionally intersect the opposite vertex (in which case it's the altitude of an isosceles triangle) but usually it will intersect exactly one of the other sides.

I've rewritten the question to put one vertex at the origin and to just ask for the perpendicular bisector of the opposite side.

The sides of the triangle are given parametrically as

$O A : \quad \quad$ $\left(x , y\right) = v \left(a , b\right) \quad \quad \quad \quad 0 \le v \le 1$

$O C : \quad \quad$ $\left(x , y\right) = w \left(c , d\right) \quad \quad \quad \quad 0 \le w \le 1$

$A C : \quad \quad$ $\left(x , y\right) = \left(a , b\right) + u \left(c - a , d - b\right) \quad \quad \quad \quad 0 \le u \le 1$

The perpendicular family to $A C$ has direction vector given by swapping coordinates and negating one; the bisector is through the midpoint of AC:

$\left(x , y\right) = \frac{1}{2} \left(a + c , b + d\right) + t \left(d - b , a - c\right) \quad \quad \quad$ for real $t$

That meets $O A$ when

$v \left(a , b\right) = \frac{1}{2} \left(a + c , b + d\right) + t \left(d - b , a - c\right)$

Two equations in two unknowns.

$v a = \frac{a + c}{2} + t \left(d - b\right)$

$v b = \frac{b + d}{2} + t \left(a - c\right)$

We're more interested in $v$ than $t$:

$2 v a \left(a - c\right) + 2 t \left(b - d\right) \left(a - c\right) = \left(a + c\right) \left(a - c\right)$

$2 v b \left(b - d\right) - 2 t \left(a - c\right) \left(b - d\right) = \left(b + d\right) \left(b - d\right)$

$v = \frac{1}{2} \cdot \frac{\left({a}^{2} + {b}^{2}\right) - \left({c}^{2} + {d}^{2}\right)}{{a}^{2} + {b}^{2} - \left(a c + b d\right)}$

The roles aren't quite symmetrical so we repeat the process for the meet of the bisector and $O C$.

$w \left(c , d\right) = \frac{1}{2} \left(a + c , b + d\right) + t \left(d - b , a - c\right)$

$2 w c \left(a - c\right) = \left(a + c\right) \left(a - c\right) - 2 t \left(b - d\right) \left(a - c\right)$

$2 w d \left(b - d\right) = \left(b + d\right) \left(b - d\right) + 2 t \left(a - c\right) \left(b - d\right)$

$w = \frac{1}{2} \cdot \frac{\left({a}^{2} + {b}^{2}\right) - \left({c}^{2} + {d}^{2}\right)}{\left(a c + b d\right) - \left({c}^{2} + {d}^{2}\right)}$

Yes I know this is getting long Socratic. It's an involved problem.

The numerators are the same in $v$ and $w$. We need to show that either $v = w = 0$ (an isosceles triangle where the perpendicular bisector is an altitude, here to the origin) or exactly one of $0 < v < 1$ or $0 < w < 1.$

~~~~~~~

We want to solve $v > 0$

$\frac{\left({a}^{2} + {b}^{2}\right) - \left({c}^{2} + {d}^{2}\right)}{{a}^{2} + {b}^{2} - \left(a c + b d\right)} > 0$

Let's first assume ${a}^{2} + {b}^{2} > {c}^{2} + {d}^{2} ,$ positive numerator, so we need a positive denominator.

${a}^{2} + {b}^{2} > a c + b d$

$| A {|}^{2} > A \cdot C = | A | \setminus | C | \setminus \cos A O C$

$| A | > | C | \setminus \cos A O C$

We're assuming $| A | > | C |$ and the cosine is never bigger than one so we've shown $| A | > | C | \implies v > 0.$

What about when ${a}^{2} + {b}^{2} < {c}^{2} + {d}^{2} ,$ or $| A | < | C |$? For $v > 0$ we need a negative denominator.

${a}^{2} + {b}^{2} < a c + b d$

$| A {|}^{2} < A \cdot C = | A | \setminus | C | \setminus \cos A O C$

$| A | < | C | \setminus \cos A O C$

$\cos A O C > | A \frac{|}{|} C |$

How about $w > 0$?

$w = \frac{1}{2} \cdot \frac{| A {|}^{2} - | C {|}^{2}}{A \cdot C - | C {|}^{2}}$

For $| A | > | C |$ we need $A \cdot C > | C {|}^{2}$ or

$| A | \setminus | C | \setminus \cos A O C > | C {|}^{2}$

$\cos A O C > | C \frac{|}{|} A |$

Unless $| A | = | C |$ exactly one of those fractions $| C \frac{|}{|} A | \mathmr{and} | A \frac{|}{|} C |$ is in cosine range, between -1 and 1.

Still not there.

~~~~~~~~~

We need to show the denominators have opposite signs.

$\left(\left({a}^{2} + {b}^{2}\right) - \left(a c + b d\right)\right) \left(\left(a c + b d\right) - \left({c}^{2} + {d}^{2}\right)\right)$

$= \left({a}^{2} + {b}^{2} + {c}^{2} + {d}^{2}\right) \left(a c + b d\right) + {\left(a c + b d\right)}^{2} + \left({a}^{2} + {b}^{2} - {c}^{2} - {d}^{2}\right) \left(a c + b d\right)$

= (ac+bd)( a^2+b^2 + c^2 + d^2 + (ac+bd) + a^2+b^2-c^2-d^2)  =(ac+bd)( 2a^2 + 2b^2 + (ac+bd) )

~ ~~~~~~

The triangle inequality says

$| A C | > | O A | + | O C |$

$| A C {|}^{2} > | O A {|}^{2} + | O C {|}^{2} + 2 | O A | \setminus | O C |$

${\left(a - c\right)}^{2} + {\left(b - d\right)}^{2} > {a}^{2} + {b}^{2} + {c}^{2} + {d}^{2} + 2 \sqrt{\left({a}^{2} + {b}^{2}\right) \left({c}^{2} + {d}^{2}\right)}$

${\left(a - c\right)}^{2} + {\left(b - d\right)}^{2} - \left({a}^{2} + {b}^{2} + {c}^{2} + {d}^{2}\right) > 2 \sqrt{\left({a}^{2} + {b}^{2}\right) \left({c}^{2} + {d}^{2}\right)}$

$- 2 a c - 2 b d > 2 \sqrt{\left({a}^{2} + {b}^{2}\right) \left({c}^{2} + {d}^{2}\right)}$

$a c - b d > 2 \sqrt{\left({a}^{2} + {b}^{2}\right) \left({c}^{2} + {d}^{2}\right)}$

The dot product and the sum of squares have a close relationship. We expand $| A C {|}^{2} :$

$| A C {|}^{2} = {\left(a - c\right)}^{2} + {\left(b - d\right)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + {d}^{2} - 2 \left(a c + b d\right)$

$\left(a c + b d\right) - \left({c}^{2} + {d}^{2}\right) = \left({a}^{2} + {b}^{2}\right) - \left(a c + b d\right) - | A C {|}^{2}$

$w = \frac{1}{2} \cdot \frac{| A {|}^{2} + | C {|}^{2}}{| A {|}^{2} - A \cdot C - | A C {|}^{2}}$

$v = \frac{1}{2} \cdot \frac{| A {|}^{2} + | C {|}^{2}}{| A {|}^{2} - A \cdot C}$

We still haven't shown we pass through at most one side or the vertex. It remains to show $- 1 < v < 1$ which ensures that the bisector passes through exactly one side.

Still more to do.

1

A triangle has corners at (9 ,7 ), (2 ,1 ), and (5 ,4 ). What is the area of the triangle's circumscribed circle?

Hammer
Featured 2 days ago

The area of the circle is aproximately $301.866$, with exact form $2210 \pi \text{/} 23$.

Explanation:

First things first, let us visualise the points we have:

These points may look collinear, but they are not.
Now, let's draw the triangle:

And finally, the circle:

We can finally start applying formulae and such.

The area of a circle, denoted $S$ in this answer, is $\pi {r}^{2}$ where $r$ is the radius.

In our case, the radius we search is the radius of the circumscribed triangle, $R$.

But how do we find $R$? Well, we could apply the Law of Sines:

$\frac{a}{\sin} A = \frac{b}{\sin} B = \frac{c}{\sin} C = 2 R$

So we must find one of the sides and the sine of its opposite angle, say $c = A B$ and $\angle C$.

The line $A B$ is defined by the points $B \left(2 , 1\right)$ and $A \left(9 , 7\right)$. Thus, the measure of $A B$ is:

$A B = \sqrt{{\left(2 - 9\right)}^{2} + {\left(1 - 7\right)}^{2}} = \sqrt{49 + 36} = \sqrt{85} = c$

So we have found $c$. To find $C$, we must use the Law of cosine, which claims that, in any triangle:

${c}^{\textcolor{red}{2}} = {a}^{\textcolor{red}{2}} + {b}^{\textcolor{red}{2}} - 2 a b \cos C$

and its analogues. We are doing this to find the value of $\cos C$, from which we can derive $\sin C$.

We got to find $a$ and $b$ too, now:

$B C = \sqrt{{\left(2 - 5\right)}^{2} + {\left(1 - 4\right)}^{2}} = \sqrt{18} = 3 \sqrt{2} = a$
$A C = \sqrt{{\left(9 - 7\right)}^{2} + {\left(7 - 4\right)}^{2}} = \sqrt{13} = b$

$\therefore {\left(\sqrt{85}\right)}^{\textcolor{red}{2}} = {\left(\sqrt{18}\right)}^{\textcolor{red}{2}} + {\left(\sqrt{13}\right)}^{\textcolor{red}{2}} - 2 \sqrt{18} \sqrt{13} \cos C$

$85 = 18 + 13 - 2 \sqrt{234} \cos C \implies \cos C = - \frac{27}{2 \sqrt{234}}$

$\textcolor{b l u e}{\cos C = - 9 \text{/} \sqrt{104}}$

We can see that $\cos C$ is negative, meaning that $C$ is an obtuse angle, as visible in the diagram above.

From the Fundamental property of Trigonometry, we have:

${\sin}^{\textcolor{red}{2}} C + {\cos}^{\textcolor{red}{2}} C = 1$
$\implies \sin C = \textcolor{red}{+} \sqrt{1 - {\cos}^{\textcolor{red}{2}} C}$

If you're wondering why $\sin C$ has to be strictly positive; it's because the sine function is positive in both the $\text{I"^"st}$ and $\text{II"^"nd}$ Quadrants.

$\sin C = \sqrt{1 - \frac{81}{104}} = \sqrt{\frac{23}{104}}$

Finally, we can find $R$ by the Law of Sines:

$\frac{c}{\sin} C = 2 R$

$\frac{\sqrt{85}}{\sqrt{\frac{23}{104}}} = 2 R \implies R = \frac{1}{2} \sqrt{\frac{8840}{23}} = \sqrt{\frac{2210}{23}}$

color(blue)(R = sqrt(2210/23)

So the area of the circle, $S$, is equal to

color(red)(S = piR^2 = 2210/23 pi ~~301.866

3

If ASB is a focal chord of a parabola such that AS=2 and SB =4, Then find the latus rectum?

dk_ch
Featured yesterday

Let the equation of the parabola in standard form be ${y}^{2} = 4 a x$.So the coordinates of its focus S will be $\left(a , 0\right)$.

Let the coordinates of one end point A of the given focal chord AB be $\left(a {t}_{1}^{2} , 2 a {t}_{1}\right)$ and that of other end point B be $\left(a {t}_{2}^{2} , 2 a {t}_{2}\right)$

AS and SB line segments are on same straight line. So their slopes must be same.

Hence $\frac{2 a {t}_{1} - 0}{a {t}_{1}^{2} - a} = \frac{2 a {t}_{2} - 0}{a {t}_{2}^{2} - a}$

$\implies {t}_{1} / \left({t}_{1}^{2} - 1\right) = {t}_{2} / \left({t}_{2}^{2} - 1\right)$

$\implies {t}_{1} \left({t}_{2}^{2} - 1\right) = {t}_{2} \left({t}_{1}^{2} - 1\right)$

$\implies {t}_{1} \left({t}_{2}^{2} - 1\right) - {t}_{2} \left({t}_{1}^{2} - 1\right) = 0$

$\implies {t}_{1} {t}_{2}^{2} - {t}_{1} - {t}_{2} {t}_{1}^{2} + {t}_{2} = 0$

$\implies {t}_{1} {t}_{2} \left({t}_{2} - {t}_{1}\right) + \left({t}_{2} - {t}_{1}\right) = 0$

$\implies \left({t}_{1} {t}_{2} + 1\right) \left({t}_{2} - {t}_{1}\right) = 0$

As A and B are different points then ${t}_{1} \ne {t}_{2}$

So ${t}_{2} = - \frac{1}{t} _ 1$

Now it is given that $A S = 2 \mathmr{and} S B = 4$

So

$\frac{S B}{S A} = \frac{4}{2} = 2$

$\implies {\left(S B\right)}^{2} / {\left(S A\right)}^{2} = 4$

$\implies \frac{{\left(a {t}_{2}^{2} - a\right)}^{2} + {\left(2 a {t}_{2} - 0\right)}^{2}}{{\left(a {t}_{1}^{2} - a\right)}^{2} + {\left(2 a {t}_{1} - 0\right)}^{2}} = 4$

$\implies \frac{{\left({t}_{2}^{2} - 1\right)}^{2} + 4 {t}_{2}^{2}}{{\left({t}_{1}^{2} - 1\right)}^{2} + 4 {t}_{1}^{2}} = 4$

$\implies {\left({t}_{2}^{2} + 1\right)}^{2} / {\left({t}_{1}^{2} + 1\right)}^{2} = 4$

$\implies \frac{{t}_{2}^{2} + 1}{{t}_{1}^{2} + 1} = 2$

Inserting ${t}_{2} = - \frac{1}{t} _ 1$ we get

$\implies \frac{\frac{1}{t} _ {1}^{2} + 1}{{t}_{1}^{2} + 1} = 2$

$\implies \frac{1}{t} _ {1}^{2} = 2$

$\implies {t}_{1}^{2} = \frac{1}{2}$

Now

$S {A}^{2} = 4$

$\implies {\left(a {t}_{1}^{2} - a\right)}^{2} + {\left(2 a {t}_{1} - 0\right)}^{2} = 4$

=>a^2[t_1^2-1)^2+4t_1^2]=4

$\implies {a}^{2} \left[{\left({t}_{1}^{2} + 1\right)}^{2}\right] = 4$

$\implies a \left({t}_{1}^{2} + 1\right) = 2$

Inserting ${t}_{1}^{2} = \frac{1}{2}$ we get

$a \left(\frac{1}{2} + 1\right) = 2$

$\implies a \times \frac{3}{2} = 2$

$\implies a = \frac{4}{3}$

So length of the Latus rectum will be

$= 4 a = 4 \cdot \frac{4}{3} = \frac{16}{3}$

A shortcut

As $S A : S B = 1 : 2$ and Coordinates of $S$ is $\left(a , 0\right)$

So
$\frac{2 a {t}_{1} \cdot 2 + 2 a {t}_{2} \cdot 1}{3} = 0$

$\implies {t}_{2} = - 2 {t}_{1}$

$\implies - \frac{1}{t} _ 1 = - 2 {t}_{1}$

$\implies {t}_{1}^{2} = \frac{1}{2}$