length of $SP$ in parabola $y^2=4x$ where $S$ is a focus and $P$ is an external point

Question

If the length of tangents drawn from point $P$ to the parabola $y^2=4x$ with focus $S$ such that the length of tangents are $\sqrt{5}$ and $\sqrt{10}$ unit respectively.Then length of $SP$ equals

What i try

Let $Q(t^2_{1},2t_{1})$ and $R(t^2_{2},2t_{2})$ be two point on the parabola where tangent from meet $P(t_{1}t_{2},t_{1}+t_{2})$ meet with parabola.

Which is nothing but point of intersection of tangent at $Q$ and $R$

$t_{1}y=x+t^2_{1}\cdots\cdots (1)$

$t_{2}y=x+t^2_{2}\cdots\cdots (2)$

Solving these two we get $x=t_{1}t_{2},t_{1}+t_{2}$ which is coordinate of $P$

Coordinate of focus $S(1,0)$

So $$\displaystyle SP=\sqrt{(1-t_{1}t_{2})^2+(t_{1}+t_{2})^2}$$

$$SP=\sqrt{(1+t_{1}^2)(1+t_{2}^2)}$$

And $$\displaystyle PQ=\sqrt{(t_{1}t_{2}-t_{1}^2)^2+(t_{1}+t_{2}-2t_{1})^2}=(t_{2}-t_{1})\sqrt{1+t_{1}^2}=\sqrt{5}\cdots (3)$$

And $$\displaystyle PR=\sqrt{(t_{1}t_{2}-t_{2}^2)^2+(t_{1}+t_{2}-2t_{2})^2}=(t_{2}-t_{1})\sqrt{1+t_{2}^2}=\sqrt{10}\cdots (4)$$

How do i find value of $t_{1},t_{2}$ in easy way or any short way to find value of $SP$

Help me please

Answer 1

Eliminating $t_2$ from your equations one obtains a quartic equation for $z=t_1^2$: $$ z^4+4z^3-24z^2-36z+16=0. $$ This has four real solutions, of which only two are positive: one of them is easy ($z=4$) but the other one is given by an involved expression (according to Mathematica): $$ z=-\frac{8}{3}+\frac{4/3\ 10^{2/3}}{\sqrt[3]{-17+3 i\sqrt{39}}}+ \frac{1}{3}\sqrt[3]{10 \left(-17+3 i \sqrt{39}\right)} \approx 0.362579. $$ Both solutions are fine but at the moment I don't see a way to avoid solving a quartic equation.

Answer 2

Let us solve with bare hands the obtained system of two algebraic equations: $$ \left\{ \begin{aligned} 5 &= (1+t^2)(s-t)^2\ ,\\ 10 &= (1+s^2)(s-t)^2\ , \end{aligned} \right. $$ where i am using $s,t$ (instead of $t_1,t_2$) to have an easier typing. Let $s,t$ be a solution (over reals) of the given system. Then $2=10/5=(1+s^2)/(1+t^2)$, which allows us to express $s^2$ in terms of $t^2$, $s^2=2t^2+1$. We substitute this in the first equation, and try to isolate $st$, the only term still involving $s$ after all. Then we take the square again to get rid of $s$. It turns now out that the substitution $X=t^2+1$ becomes handy, so for instance $s^2=2t^2+1=2X-1$, and we obtain an equation of degree four in $X$. All in all: $$ \begin{aligned} 5 &= (1+t^2)\Big(\ (2t^2+1)-2st+t^2\ \Big)\ ,\\ \frac5{1+t^2}-(3t^2+1)&=-2st\ ,\\ \left(\frac5{1+t^2}-(3t^2+1)\right)^2&=4s^2t^2\ ,\\ \left(\frac5X-(3X-2)\right)^2&=4(2X-1)(X-1)\ ,\\ 25-10X(3X-2)+(3X-2)^2X^2 &=4X^2(2X-1)(X-1)\ ,\\ X^4 - 30X^2 + 20X +25 &=0\ ,\\ (X-5)(X^3 + 5X^2 -5X - 5) &=0\ . \end{aligned} $$ The solution $X=5$ corresponds to $t^2=4$, so $t=\pm2$, $s=\pm 3$, and we accept only the pairs $(s,t)$ with $5=(1+t^2)(s-t)^2$ which are $(s,t)\in\{\ (2,3)\ ,\ (-2,-3)\ \}$. This leads to the points $P_\pm$ of the form $(6,\pm 5)$ mirrored by the symmetry axis of the parabole, and they turn out to be solutions, both have distance $\sqrt{(6-1)^2+5^2}=5\sqrt 2$ from the focal point $S=(1,0)$.

Furthermore, the polynomial $(X^3 + 5X^2 -5X - 5)$ has exactly one positive solution, $1.362578565668247\dots$ - and it leads to other two mirrored points
$P_\pm$ - however computations are not for pen and paper. Below, aided by a computer algebra system (CAS), sage, one can obtain the minimal polynomial for $s$, for $t$, and also for the distance from $P(st,s+t)$ to $S=(1,0)$.

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Computer aided solution:

In sage, the two given algebraic equations in $s,t$ are defining an ideal, and the variety associated to the ideal is the set of all points satisfying the given equations. We can ask for points with both coordinates in some specific field. The field of interest for us is $\Bbb Q=$QQbar, the algebraic closure of $\Bbb Q$. It contains as subfield the field $\Bbb A=$AA of real algebraic numbers. We ask for the points in this variety, they are finitely many, and for each such point we ask for the minimal polynomial of the expression $$(st-1)^2+(s+t)^2\ ,$$ which is the squared distance from the focal point $S=(1,0)$ to a solution point $P(st,s+t)$ for our problem. Note that computations are done exaclty, each point in QQbar comes with a minimal polynomial in $\Bbb Q[x]$, an exact ring, together with a (good) approximation that separates the roots of this minimal polynomial.

Code:

R.<s,t> = PolynomialRing(QQ)
J = R.ideal([ 5 - (1 + t^2)*(s - t)^2, 10 - (1 + s^2)*(s - t)^2])
print(f'Elimination of s gives:\n{J.elimination_ideal(s).gens()[0].factor()} = 0')
print(f'Elimination of t gives:\n{J.elimination_ideal(t).gens()[0].factor()} = 0')
print('\nThe variety defined by J has the points over AA:')
V = J.variety(ring=AA)
for v in V:
   print(f'v = {v} with components:')
   vs, vt = v[s], v[t]
   print(f's = {vs} with minimal polynomial {vs.minpoly()}')
   print(f't = {vt} with minimal polynomial {vt.minpoly()}')
   SP2 = (vs*vt - 1)^2 + (vs + vt)^2
   print(f'Then SP² = {SP2} has minimal polynomial {SP2.minpoly()}\n')

And we obtain the following information:

Elimination of s gives:
(t - 2) * (t + 2) * (t^6 + 8*t^4 + 8*t^2 - 4) = 0
Elimination of t gives:
(s - 3) * (s + 3) * (s^6 + 13*s^4 + 3*s^2 - 49) = 0
The variety defined by J has the points over AA:
v = {t: 2, s: 3} with components:
s = 3 with minimal polynomial x - 3
t = 2 with minimal polynomial x - 2
Then SP² = 50 has minimal polynomial x - 50
v = {t: -2, s: -3} with components:
s = -3 with minimal polynomial x + 3
t = -2 with minimal polynomial x + 2
Then SP² = 50 has minimal polynomial x - 50
v = {t: -0.6021449706409969?, s: 1.313452371171675?} with components:
s = 1.313452371171675? with minimal polynomial x^6 + 13x^4 + 3x^2 - 49
t = -0.6021449706409969? with minimal polynomial x^6 + 8x^4 + 8x^2 - 4
Then SP² = 3.713240695237075? has minimal polynomial x^3 - 70x^2 + 300x - 200
v = {t: 0.6021449706409969?, s: -1.313452371171675?} with components:
s = -1.313452371171675? with minimal polynomial x^6 + 13x^4 + 3x^2 - 49
t = 0.6021449706409969? with minimal polynomial x^6 + 8x^4 + 8x^2 - 4
Then SP² = 3.713240695237075? has minimal polynomial x^3 - 70x^2 + 300x - 200

The points $\pm(2,3)$ are in the list, and the corresponding squared distance is $SP^2=50$, so $SP=\sqrt{50}=5\sqrt 2$, we did this also manually.

But there are two other mirrored points, they are leading to a solution with squared distance $SP^2$ which is the root of $$ x^3 - 70x^2 + 300x - 200 $$ near $3.713240695237075\dots$ - and the explicit formula for this root is: $$ \begin{aligned} SP^2 &= \frac {70}3 + \frac 43\sqrt[3]{25(157+3\sqrt{-39})} + \frac 43\sqrt[3]{25(157-3\sqrt{-39})} \\ &= \frac {70}3 + \frac 83\Re\sqrt[3]{25(157+3\sqrt{-39})} \ , \end{aligned} $$ where above, in the first expression the cubic roots of $$ 25(157\pm3\sqrt{-39}) $$ should be taken to be complex conjugated, i.e. their product has to be $25^2(157^2+3^2\cdot 9) = 15625000= 2^3\cdot 5^9=(2\cdot 5^3)^3$. Also, among the three cubic roots one should pick the one near $-7.3575347392860\dots - 13.9952378529341\dots i$ .

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A picture of the solutions (one solution taken from each mirrored pair, so that the solution points are not to close to each other) is as follows:

The distance $SP_1$ above is given by geogebra approximatively as

which matches

sage: SP2
3.713240695237075?
sage: sqrt(SP2)
1.926977087366914?