Consider the ellipse E given by $\frac{x^2}{18}+\frac{y^2}{12}=1$. Let H be the hyperbola whose eccentricity is the reciprocal of the eccentricity of E and whose foci are the same as that of E. Let P and Q be the points of intersection of H and the parabola $x^2=\sqrt{5}y$ in the first quadrant. Let d be the distance between P and Q. If a and b are the integers such that $d^2=a+b\sqrt{5}$, then the value of a − b is .

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The correct answer is 18.

Step 1: Determine the eccentricity and foci of the ellipse E

The given equation of the ellipse E is:

$\frac{x^2}{18}+\frac{y^2}{12}=1$

Here, $a_E^2=18$ and $b_E^2=12$.

The eccentricity $e_E$ of the ellipse is calculated as:

$e_E=\sqrt{1-\frac{b_E^2}{a_E^2}}=\sqrt{1-\frac{12}{18}}=\sqrt{1-\frac{2}{3}}=\frac{1}{\sqrt{3}}$

The distance of the foci from the center is given by $c=a_E{e}_E$:

$c=\sqrt{18}·\frac{1}{\sqrt{3}}=\sqrt{6}$

Thus, the foci of the ellipse are located at $(\pm \sqrt{6},0)$.

Step 2: Find the equation of the hyperbola H

The eccentricity of the hyperbola H is the reciprocal of the eccentricity of E:

$e_H=\frac{1}{e_E}=\sqrt{3}$

Since H shares the same foci as E, its foci are also at $(\pm \sqrt{6},0)$. For a standard hyperbola $\frac{x^2}{a_H^2}-\frac{y^2}{b_H^2}=1$, the focus distance is $a_H{e}_H=\sqrt{6}$:

$a_H\sqrt{3}=\sqrt{6}\Rightarrow a_H=\sqrt{2}\Rightarrow a_H^2=2$

Using the relationship $b_H^2=a_H^2(e_H^2-1)$:

$b_H^2=2(3-1)=4$

Therefore, the equation of the hyperbola H is:

$\frac{x^2}{2}-\frac{y^2}{4}=1\Rightarrow 2x^2-y^2=4$

Step 3: Find the points of intersection P and Q

We solve for the points of intersection between the hyperbola $2x^2-y^2=4$ and the parabola $x^2=\sqrt{5}y$ in the first quadrant.

Substituting $x^2=\sqrt{5}y$ into the hyperbola equation:

$2\left(\sqrt{5}y\right)-y^2=4\Rightarrow y^2-2\sqrt{5}y+4=0$

Solving this quadratic equation for $y$ using the quadratic formula:

$y=\frac{2\sqrt{5}\pm \sqrt{{\left(2\sqrt{5}\right)}^2-4\left(1\right)\left(4\right)}}{2}=\frac{2\sqrt{5}\pm \sqrt{20-16}}{2}=\sqrt{5}\pm 1$

Both values of $y$ are positive, corresponding to two distinct points in the first quadrant:

For $y_1=\sqrt{5}+1$:

$x_1^2=\sqrt{5}(\sqrt{5}+1)=5+\sqrt{5}\Rightarrow x_1=\sqrt{5+\sqrt{5}}$

For $y_2=\sqrt{5}-1$:

$x_2^2=\sqrt{5}(\sqrt{5}-1)=5-\sqrt{5}\Rightarrow x_2=\sqrt{5-\sqrt{5}}$

Thus, the coordinates of points P and Q are:

$P=(\sqrt{5+\sqrt{5}},\sqrt{5}+1)$ and $Q=(\sqrt{5-\sqrt{5}},\sqrt{5}-1)$

Step 4: Calculate the distance squared $d^2$ between P and Q

Using the distance formula:

$d^2={(x_1-x_2)}^2+{(y_1-y_2)}^2$

First, evaluate ${(y_1-y_2)}^2$:

$y_1-y_2=(\sqrt{5}+1)-(\sqrt{5}-1)=2\Rightarrow {(y_1-y_2)}^2=4$

Next, evaluate ${(x_1-x_2)}^2=x_1^2+x_2^2-2x_1{x}_2$:

$x_1^2+x_2^2=(5+\sqrt{5})+(5-\sqrt{5})=10$

$x_1{x}_2=\sqrt{(5+\sqrt{5})(5-\sqrt{5})}=\sqrt{25-5}=\sqrt{20}=2\sqrt{5}$

${(x_1-x_2)}^2=10-2\left(2\sqrt{5}\right)=10-4\sqrt{5}$

Combining both terms:

$d^2=(10-4\sqrt{5})+4=14-4\sqrt{5}$

Step 5: Find the value of a − b

We are given that $d^2=a+b\sqrt{5}$. Comparing this with $d^2=14-4\sqrt{5}$, we get:

$a=14$ and $b=-4$

Now, calculate $a-b$:

$a-b=14-(-4)=14+4=18$