Focuses of the hyperbola F1 (4,2), F2 (−1, −10) and the tangent equation 3x + 4y - 5 = 0 are given. Find the semiaxes of the hyperbola.
I’ve tried using the definition of hyperbola and the equation for the tangent line to the quadratic curve to form some sort of system of equations, but it didn’t work out.
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The product of distances from focis to any tangent line is the square of conjugate semiaxis.
The distances from $F_1$ and $F_2$ to the tangent line are respectfully $15/5$ and $48/5$, so $b=12/\sqrt{5}$.
The distance from $F_1$ to $F_2$ is equal to $2c$, then $4c^2=13^2$.
Then $a= \sqrt{c^2-b^2} = \sqrt{269}/(2\sqrt5)$.
$\sqrt{(x-4)^2+(y-2)^2}-\sqrt{(x+1)^2+(y+10)^2}=\pm 2a$
$\sqrt{(x-4)^2+(y-2)^2}=\pm 2a +\sqrt{(x+1)^2+(y+10)^2}$
$(x-4)^2+(y-2)^2=4a^2 \pm 4a\sqrt{(x+1)^2+(y+10)^2}+(x+1)^2+(y+10)^2$
$((x-4)^2+(y-2)^2-4a^2 -((x+1)^2+(y+10)^2))^2=16a^2((x+1)^2+(y+10)^2)$
$(-4 a^2 - 10 x - 24 y - 81)^2=16a^2((x+1)^2+(y+10)^2)$
$16 a^4 - 16 a^2 x^2 + 48 a^2 x - 16 a^2 y^2 - 128 a^2 y - 968 a^2 + 100 x^2 + 480 x y + 1620 x + 576 y^2 + 3888 y + 6561=0$
Dual:
$-16a^2(2a-13)(2a+13)(4a^2y^2-89y^2+168xy+32y+4a^2x^2-153x^2-12x-4)=0$
The tangent is $-\frac35x-\frac45y+1=0$ so setting $(-\frac35,-\frac45)$ into the dual we get $a^2=\frac{269}{20}.$
That is the equation is $$2255y^2+3000xy+13540y-720x^2+14160x-22276=0$$ or $$2255(y+4)^2+3000(x-\frac32)(y+4)-720(x-\frac32)^2-38736=0$$
The matrix has eigenvalues $2880,-1345,$ so the canonical form is $$(20y'^2)/269-(5x'^2)/144=1,$$ so the semi-axes are $\sqrt{\frac{269}{20}}, \frac{12}{\sqrt{5}}.$