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Step-by-Step Solution
Step 1: Write down the given ellipse and the point of tangency
The ellipse is given by
x^2 + 4y^2 = 5 ,
and the point of tangency on this ellipse is P(1,\,1).
Step 2: Find the equation of the tangent at the given point
One way to find the tangent to F(x,\,y)=0 at (x_0,\,y_0) is to use
\displaystyle F_x(x_0,\,y_0)\,(x - x_0)\;+\;F_y(x_0,\,y_0)\,(y - y_0)=0,
where F(x,y)=x^2 + 4y^2 -5.
Compute partial derivatives:
F_x(x,y) = 2x.
F_y(x,y) = 8y.
At the point (1,\,1): F_x(1,1)=2,\; F_y(1,1)=8. Hence, the tangent is
2\,(x - 1)\;+\;8\,(y - 1)\;=\;0 \;\;\Longrightarrow\;\;2x -2 + 8y -8 = 0,
which simplifies to
\displaystyle x\;+\;4y\;=\;5.
Step 3: Identify the region whose area is to be found
We want the area of the region bounded by:
The ellipse x^2 + 4y^2 = 5,
The tangent line T: x + 4y = 5,
The vertical lines x=1 and x=\sqrt{5}.
Geometrically, for x between 1 and \sqrt{5}, the top boundary of the ellipse is
y = \frac{1}{2}\sqrt{5 - x^2}
(since 4y^2 = 5 - x^2 \implies y^2 = \frac{5 - x^2}{4} ),
and the line is y = \frac{5 - x}{4}.
Step 4: Set up the definite integral for the desired area
For each x from 1 to \sqrt{5}, the vertical “strip” in this region has height
\bigl[\tfrac{5 - x}{4}\bigr] - \bigl[\tfrac{1}{2}\sqrt{5 - x^2}\bigr].
Thus, the area I is
\displaystyle
I \;=\;\int_{x=1}^{x=\sqrt{5}}
\left[
\frac{5 - x}{4}\;-\;\frac{1}{2}\,\sqrt{5 - x^2}
\right]
\,dx.
Step 5: Split the integral into two simpler integrals
\[
I \;=\;\underbrace{\int_{1}^{\sqrt{5}} \frac{5 - x}{4} \,dx}_{I_1}
\;-\;
\underbrace{\int_{1}^{\sqrt{5}} \frac{1}{2}\,\sqrt{5 - x^2}\,dx}_{I_2}.
\]
(a) Compute I_1
\[
I_1
\;=\;
\int_{1}^{\sqrt{5}} \frac{5 - x}{4}\,dx
\;=\;
\frac{1}{4}\,\int_{1}^{\sqrt{5}} (5 - x)\,dx.
\]
Inside the integral,
\[
\int (5 - x)\,dx \;=\;5\,x - \tfrac{x^2}{2}.
\]
Hence,
\[
I_1
=
\frac{1}{4}\,\Bigl[\,
5x - \tfrac{x^2}{2}
\Bigr]_{1}^{\sqrt{5}}
=
\frac{1}{4}
\Bigl(
\bigl[5(\sqrt{5}) - \tfrac{(\sqrt{5})^2}{2}\bigr]
-
\bigl[5(1) - \tfrac{1^2}{2}\bigr]
\Bigr).
\]
Evaluate step by step:
At x = \sqrt{5}\!: 5(\sqrt{5}) - \tfrac{5}{2} = 5\sqrt{5} - 2.5.
At x = 1\!: 5 - \tfrac{1}{2} = 4.5.
Difference: (5\sqrt{5} - 2.5)\;-\;4.5 = 5\sqrt{5} - 7.
Thus,
\[
I_1
=
\frac{1}{4}\,(\,5\sqrt{5} - 7\,)
=
\frac{5\sqrt{5}}{4} \;-\;\frac{7}{4}.
\]
(b) Compute I_2
\[
I_2
=
\int_{1}^{\sqrt{5}} \frac{1}{2}\,\sqrt{5 - x^2}\,dx
=
\frac{1}{2}
\int_{1}^{\sqrt{5}} \sqrt{5 - x^2}\,dx.
\]
Use the trigonometric substitution x = \sqrt{5}\,\sin\theta, so that
dx = \sqrt{5}\,\cos\theta \,d\theta and \sqrt{5 - x^2} = \sqrt{5}\,\cos\theta.
Under this substitution:
When x = 1, we get 1 = \sqrt{5}\,\sin\theta \implies \sin\theta = \tfrac{1}{\sqrt{5}} \implies \theta = \sin^{-1}\bigl(\tfrac{1}{\sqrt{5}}\bigr).
When x = \sqrt{5}, we get \sin\theta = 1 \implies \theta = \tfrac{\pi}{2}.
Hence,
\[
I_2
=
\frac{1}{2}\;
\int_{\sin^{-1}(1/\sqrt{5})}^{\pi/2}
\sqrt{5}\,\cos\theta \;\times\;\sqrt{5}\,\cos\theta\,d\theta
=
\frac{1}{2}
\int_{\sin^{-1}(1/\sqrt{5})}^{\pi/2}
5\,\cos^2\theta
\,d\theta
=
\frac{5}{2}
\int_{\sin^{-1}(1/\sqrt{5})}^{\pi/2}
\cos^2\theta
\,d\theta.
\]
A standard result is
\displaystyle \int \cos^2\theta\,d\theta \;=\;\frac{\theta}{2} + \tfrac{\sin(2\theta)}{4}.
Thus,
\[
I_2
=
\frac{5}{2}\,
\left[
\frac{\theta}{2} + \frac{\sin(2\theta)}{4}
\right]_{\theta=\sin^{-1}(1/\sqrt{5})}^{\,\theta=\tfrac{\pi}{2}}.
\]
Let
\[
J(\theta) \;=\; \frac{5}{2}\,\left(\frac{\theta}{2} + \frac{\sin(2\theta)}{4}\right)
\;=\;
\frac{5\theta}{4} + \frac{5}{8}\,\sin(2\theta).
\]
Therefore,
\[
I_2 = J\Bigl(\tfrac{\pi}{2}\Bigr) - J\Bigl(\sin^{-1}(\tfrac{1}{\sqrt{5}})\Bigr).
\]
At \theta = \tfrac{\pi}{2}\!: \sin(2\cdot\tfrac{\pi}{2}) = \sin(\pi) = 0, so
J\bigl(\tfrac{\pi}{2}\bigr) = \frac{5(\pi/2)}{4} + 0 = \frac{5\pi}{8}.
Let \theta_0=\sin^{-1}\bigl(\tfrac{1}{\sqrt{5}}\bigr). Then
\sin(2\theta_0)=2\,\sin(\theta_0)\,\cos(\theta_0)=2 \cdot \tfrac{1}{\sqrt{5}} \cdot \tfrac{2}{\sqrt{5}} = \tfrac{4}{5}.
Hence
\[
J(\theta_0)
=
\frac{5\,\theta_0}{4}
+
\frac{5}{8}\,\left(\tfrac{4}{5}\right)
=
\frac{5\,\theta_0}{4} + \frac{1}{2}.
\]
Thus,
\[
I_2
=
\frac{5\pi}{8}
-
\biggl(
\frac{5\,\theta_0}{4} + \frac{1}{2}
\biggr)
\;=\;
\frac{5\pi}{8}
-\frac{5\,\theta_0}{4}
-\frac{1}{2},
\quad
\text{where }
\theta_0 = \sin^{-1}\left(\frac{1}{\sqrt{5}}\right).
\]
Step 6: Combine the two parts to get the total area
\[
I
\;=\;
I_1 \;-\; I_2
\;=\;
\Bigl(\tfrac{5\sqrt{5}}{4} - \tfrac{7}{4}\Bigr)
\;-\;
\Bigl(\tfrac{5\pi}{8} - \tfrac{5\,\theta_0}{4} - \tfrac{1}{2}\Bigr).
\]
Simplify carefully:
Collect constant terms -\tfrac{7}{4} and +\tfrac{1}{2} = +\tfrac{2}{4} gives -\tfrac{5}{4}.
So I_1 - I_2 yields
\[
I
=
\tfrac{5\sqrt{5}}{4}
-\tfrac{5}{4}
-\tfrac{5\pi}{8}
+
\tfrac{5\,\theta_0}{4},
\quad
\theta_0 = \sin^{-1}\bigl(\tfrac{1}{\sqrt{5}}\bigr).
\]
Step 7: Convert \sin^{-1} to \cos^{-1} form
We want an expression involving \cos^{-1}\!\bigl(\tfrac{1}{\sqrt{5}}\bigr).
Note that
\[
\sin^{-1}(x) = \frac{\pi}{2} - \cos^{-1}(x).
\]
Hence,
\[
\theta_0
=
\sin^{-1}\bigl(\tfrac{1}{\sqrt{5}}\bigr)
=
\frac{\pi}{2}
-
\cos^{-1}\Bigl(\tfrac{1}{\sqrt{5}}\Bigr).
\]
Substitute this back:
\[
\tfrac{5\,\theta_0}{4}
=
\tfrac{5}{4}
\Bigl(
\tfrac{\pi}{2}
-
\cos^{-1}\!\bigl(\tfrac{1}{\sqrt{5}}\bigr)
\Bigr)
=
\tfrac{5\pi}{8}
-
\tfrac{5}{4}\,\cos^{-1}\!\bigl(\tfrac{1}{\sqrt{5}}\bigr).
\]
In the expression for I, the \tfrac{5\pi}{8} and -\tfrac{5\pi}{8} cancel out, giving
\[
I
=
\frac{5\sqrt{5}}{4}
-\frac{5}{4}
-
\frac{5}{4}\,\cos^{-1}\!\Bigl(\tfrac{1}{\sqrt{5}}\Bigr).
\]
Hence the area can be written in the form
\[
\alpha\,\sqrt{5} \;+\;\beta \;+\;\gamma\,\cos^{-1}\!\Bigl(\tfrac{1}{\sqrt{5}}\Bigr),
\]
with
\[
\alpha = \frac{5}{4},
\quad
\beta = -\frac{5}{4},
\quad
\gamma = -\frac{5}{4}.
\]
Step 8: Compute the required absolute sum
\[
\alpha + \beta + \gamma
=
\frac{5}{4} \;-\;\frac{5}{4}\;-\;\frac{5}{4}
=
-\frac{5}{4}.
\]
Taking the absolute value,
\[
\bigl|\,\alpha + \beta + \gamma\bigr|
=
\frac{5}{4}.
\]
Final Answer
\displaystyle \left|\alpha + \beta + \gamma\right| \;=\; \frac{5}{4}.
Reference Image (Provided Solution)
