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Step-by-step Solution
Step 1: Write the Ellipse in Standard Form
The given ellipse is
$x^2 + a^2\,y^2 = 25\,a^2.$
Divide through by $25\,a^2$:
$ \displaystyle \frac{x^2}{25\,a^2} + \frac{y^2}{25} = 1.$
So, in standard form, the ellipse becomes
$ \displaystyle \frac{x^2}{(5\,a)^2} + \frac{y^2}{5^2} = 1, $
meaning its semi-major axis is $5\,a$ (along the $x$-axis) and semi-minor axis is $5$ (along the $y$-axis).
Step 2: Eccentricity of the Ellipse
For an ellipse in the form $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$ (with $A > B$), the eccentricity $e_1$ satisfies
$ \displaystyle e_1 = \sqrt{1 - \frac{B^2}{A^2}}. $
Here, $A = 5\,a$ and $B = 5,$ giving
$ \displaystyle e_1 = \sqrt{1 - \frac{5^2}{(5\,a)^2}}
= \sqrt{1 - \frac{25}{25\,a^2}}
= \sqrt{1 - \frac{1}{a^2}}. $
So,
$ \displaystyle e_1^2 = 1 - \frac{1}{a^2}. $
Step 3: Write the Hyperbola in Standard Form
The given hyperbola is
$ x^2 - a^2\,y^2 = 5. $
Divide by 5 to get:
$ \displaystyle \frac{x^2}{5} - \frac{a^2\,y^2}{5} = 1
\quad\Longrightarrow\quad
\frac{x^2}{5} - \frac{y^2}{\frac{5}{a^2}} = 1. $
Hence, in standard form, $A^2 = 5$ and $B^2 = \frac{5}{a^2}.$
Step 4: Eccentricity of the Hyperbola
For a hyperbola in the form $\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1,$ its eccentricity $e_2$ satisfies
$ \displaystyle e_2 = \sqrt{1 + \frac{B^2}{A^2}}. $
Here, $A^2 = 5$ and $B^2 = \frac{5}{a^2}.$ Thus,
$ \displaystyle e_2 = \sqrt{1 + \frac{5/a^2}{5}}
= \sqrt{1 + \frac{1}{a^2}}. $
So,
$ \displaystyle e_2^2 = 1 + \frac{1}{a^2}. $
Step 5: Relate the Two Eccentricities
The question states that the ellipse’s eccentricity $e_1$ is $b$ times the hyperbola’s eccentricity $e_2$:
$ \displaystyle e_1 = b\,e_2 \quad\Longrightarrow\quad e_1^2 = b^2\,e_2^2.
$
Substitute
$ e_1^2 = 1 - \frac{1}{a^2} $
and
$ e_2^2 = 1 + \frac{1}{a^2} $
to get:
$ \displaystyle 1 - \frac{1}{a^2} = b^2\Bigl(1 + \frac{1}{a^2}\Bigr).
$
Rearranging,
$ \displaystyle \frac{a^2 - 1}{a^2} = b^2 \,\frac{a^2 + 1}{a^2}. $
Simplifying,
$ \displaystyle b^2 = \frac{a^2 - 1}{a^2 + 1}.
$
Step 6: Find the Minimum Distance a Between y=e^x and y=log(x)
We interpret $a$ as the minimum distance between the curves $y=e^x$ and $y=\log_e(x).$ Notice that $y=\log_e(x)$ is the inverse function of $y=e^x,$ implying they are mirror images about the line $y=x.$ For the distance to be minimal, we look for points on each curve that share parallel tangents also parallel to $y=x.$
• The slope of the tangent to $y = e^x$ is $\frac{dy}{dx} = e^x.$
Parallel to $y=x$ means slope $=1,$ so $e^x = 1 \implies x=0.$
Then $y=e^0=1.$
Hence the point on $y=e^x$ is $(0,1).$
• The slope of the tangent to $y=\log_e(x)$ is $\frac{dy}{dx}=\frac{1}{x}.$
Parallel to $y=x$ (slope $=1$) means $\frac{1}{x}=1 \implies x=1.$
Then $y=\log_e(1)=0.$
Hence the point on $y=\log_e(x)$ is $(1,0).$
The distance between these two points is:
$ \displaystyle \sqrt{(1-0)^2 + (0-1)^2}
= \sqrt{1 + 1}
= \sqrt{2}. $
Hence, $a = \sqrt{2}.$
Step 7: Substitute a = √2 to Find b²
From
$ \displaystyle b^2 = \frac{a^2 - 1}{a^2 + 1}, $
substitute $ a^2 = 2:$
$ \displaystyle b^2 = \frac{2 - 1}{2 + 1}
= \frac{1}{3}.
$
Step 8: Evaluate the Required Expression
The expression we need is
$ \displaystyle a^2 + \frac{1}{b^2}. $
Since $a^2 = 2$ and $1/b^2 = 3,$ we get:
$ \displaystyle a^2 + \frac{1}{b^2}
= 2 + 3
= 5.
Final Answer
The value of $a^2 + \frac{1}{b^2}$ is
5.
Supporting Figure
