SAT1000 - P767

Geometry Level pending

As shown above, the hyperbola: $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1\ (a>0,b>0)$ has right focus point $F$ , right vertex $A$ .

Line $l_1$ passes through $F$ and it intersects with the hyperbola at point $B,C$ , $l_2$ passes through $B$ and $l_2 \perp AC$ , $l_3$ passes through $C$ and $l_2 \perp AB$ , $l_2, l_3$ intersects at point $D$ .

If the distance from $D$ to line $l_1$ is less than $a+\sqrt{a^2+b^2}$ , what is the range of the slope of the hyperbola's asymptotes?

Have a look at my problem set: SAT 1000 problems

(-\sqrt{2},0) \cup (0,\sqrt{2})

(-1,0) \cup (0,1)

(-\infty,-1) \cup (1,+\infty)

(-\infty,-\sqrt{2}) \cup (\sqrt{2},+\infty)

1 solution

A Former Brilliant Member
Jun 9, 2020

Position coordinates of $B$ are $(\sqrt {a^2+b^2},\frac{b^2}{a})$ .

Slope of $\overline {AC}$ is $-\dfrac{b^2}{a\sqrt {a^2+b^2}-a^2}\implies$

slope of $\overline {BD}$ is $\dfrac{a\sqrt {a^2+b^2}-a^2}{b^2}$ .

So, the equation of $\overline {BD}$ is

$y=\dfrac{a\sqrt {a^2+b^2}-a^2}{b^2}x-\dfrac{a(a^2+b^2)-a^2\sqrt {a^2+b^2}}{b^2}+\dfrac {b^2}{a}$ .

$y$ -coordinate of $D$ is zero. So it's $x$ -coordinate is

$-\dfrac {1}{a\sqrt {a^2+b^2}-a^2}\left (\dfrac {b^4-a^2(a^2+b^2)+a^3\sqrt {a^2+b^2}}{a}\right )$ .

So, by the given condition,

$\dfrac{1}{a^2\sqrt {a^2+b^2}-a^3}\left (a^3\sqrt {a^2+b^2}+b^4-a^2(a^2+b^2)\right ) <a$

$\implies b^4-a^2b^2<0\implies b^2<a^2\implies b<a$ or $b>-a$ .

Hence the slope of the asymptotes of the hyperbola lies in the range $\boxed {(-1,0)\cup (0,1)}$ .

SAT1000 - P767

1 solution

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