87 Given the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0), F_{1}$ is the left focus, $P$ is the vertex $(0, b), A B$ is a chord passing through $F_{1}$, $A$ is in the $\mathrm{I}$ quadrant (of course $A, B$ are on the ellipse). If $\angle A P B=90^{\circ}$, try to find the slope of the line $A B$.
$87 A B$ equation is $y=k(x-c)$. Let $y^{\prime}=y-b$ (i.e., translate the $x$-axis so that $P$ becomes the origin of the new coordinates), then
$$
y^{\prime}-k x=-(b+k c)
$$

The ellipse equation becomes $\frac{x^{2}}{a^{2}}+\frac{\left(y^{\prime}+b\right)^{2}}{b^{2}}=1$, i.e.,
$$
\frac{x^{2}}{a^{2}}+\frac{y^{\prime 2}+2 b y^{\prime}}{b^{2}}=0
$$

From (1) we get $1=-\frac{y^{\prime}-k x}{b+k c}$ (obviously $A B$ does not pass through $P$, so $b+k c \neq 0$), substituting into (2) we get the homogeneous equation
$$
\frac{x^{2}}{a^{2}}+\frac{y^{\prime 2}}{b^{2}}+\frac{2 b y^{\prime}}{b^{2}} \cdot\left(-\frac{y^{\prime}-k x}{b+k c}\right)=0
$$
(3) represents two lines passing through the new origin $P$. Since it passes through $A, B$ (coordinates of $A, B$ satisfy (1), (2), so they satisfy (3)), these two lines are $P A, P B$. Given $P A \perp P B$, we have
$$
\left(\frac{y^{\prime}}{x}\right)_{1} \cdot\left(\frac{y^{\prime}}{x}\right)_{2}=-1
$$
$\left(\frac{y^{\prime}}{x}\right)_{1},\left(\frac{y^{\prime}}{x}\right)_{2}$ are the two roots of (3) (as an equation of $\frac{y^{\prime}}{x}$). By Vieta's theorem, (4) is
$$
\frac{1}{a^{2}} \div\left(\frac{1}{b^{2}}-\frac{2}{b(b+k c)}\right)=-1
$$

Solving (5) we get
$$
k=\frac{b c}{a^{2}+b^{2}} .
$$