Let a and b be positive real numbers such tha | vedclass

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(A, D) The normal at $P$ makes equal intercepts on the coordinate axes,so its slope is $-1$. Thus,the slope of the tangent at $P$ is $1$.The tangent p

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$A, D$

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(A, D) The normal at $P$ makes equal intercepts on the coordinate axes,so its slope is $-1$. Thus,the slope of the tangent at $P$ is $1$.The tangent passes through $(1, 0)$ with slope $1$,so its equation is $y - 0 = 1(x - 1)$,or $x - y = 1$.The equation of the tangent at $P(x_1, y_1)$ is $\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1$. Comparing this with $x - y = 1$,we get $\frac{x_1}{a^2} = 1$ and $\frac{y_1}{b^2} = 1$,so $x_1 = a^2$ and $y_1 = b^2$.Since $P(a^2, b^2)$ lies on the hyperbola,$\frac{(a^2)^2}{a^2} - \frac{(b^2)^2}{b^2} = 1$,which simplifies to $a^2 - b^2 = 1$.The normal at $P(a^2, b^2)$ has slope $-1$,so its equation is $y - b^2 = -1(x - a^2)$,or $x + y = a^2 + b^2$.The $x$-intercept of the normal is $x = a^2 + b^2$. The tangent $x - y = 1$ has $x$-intercept $x = 1$.The triangle is formed by the tangent,the normal,and the $x$-axis. The base is the distance between the $x$-intercepts: $(a^2 + b^2) - 1 = (a^2 - 1) + b^2 = b^2 + b^2 = 2b^2$. The height is the $y$-coordinate of $P$,which is $b^2$.Thus,$\Delta = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes (2b^2) 	imes b^2 = b^4$. So $(D)$ is true.For eccentricity $e$,$e^2 = 1 + \frac{b^2}{a^2} = 1 + \frac{a^2 - 1}{a^2} = 2 - \frac{1}{a^2}$.Since $a > 1$,$0 < \frac{1}{a^2} < 1$,so $1 < 2 - \frac{1}{a^2} < 2$,which means $1 < e^2 < 2$,so $1 < e < \sqrt{2}$. Thus $(A)$ is true.

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