Question

If the perpendicular distance from the focus of an ellipse $\frac{x^2}{9} + \frac{y^2}{b^2} = 1$ ($b<3$) to its corresponding directrix is $\frac{4}{\sqrt{5}}$, then the slope of the tangent to this ellipse drawn at $(\frac{3}{\sqrt{2}}, \frac{b}{\sqrt{2}})$ is

• A. $-2/3$
• B. $2/3$
• C. $-3/2$
• D. $3/2$

Hint:

The distance from a focus to the corresponding directrix in an ellipse is $a/e - ae$. The equation of the tangent at $(x_1, y_1)$ is found by the replacement $x^2 \to xx_1$, $y^2 \to yy_1$.

Answer 1

The Correct Option is A

For the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, we are given $a^2=9$, so $a=3$.
The coordinates of a focus are $(ae, 0)$ and the equation of the corresponding directrix is $x = a/e$. 
The distance from the focus to the directrix is $\frac{a}{e} - ae$. 
We are given this distance is $\frac{4}{\sqrt{5}}$. 
$\frac{a}{e} - ae = a(\frac{1}{e} - e) = a(\frac{1-e^2}{e}) = \frac{4}{\sqrt{5}}$. 
We also know that for an ellipse, $b^2 = a^2(1-e^2)$, so $1-e^2 = \frac{b^2}{a^2}$. 
Substitute this into the distance equation: $a\left(\frac{b^2/a^2}{e}\right) = \frac{b^2}{ae} = \frac{4}{\sqrt{5}}$. 
Also, from $b^2=a^2(1-e^2) \implies e^2 = 1 - \frac{b^2}{a^2} = \frac{a^2-b^2}{a^2} \implies e = \frac{\sqrt{a^2-b^2}}{a}$. 
Substitute $a=3$: $e = \frac{\sqrt{9-b^2}}{3}$. 
Now substitute this into $\frac{b^2}{ae} = \frac{4}{\sqrt{5}}$: 
$\frac{b^2}{3 \cdot \frac{\sqrt{9-b^2}}{3}} = \frac{b^2}{\sqrt{9-b^2}} = \frac{4}{\sqrt{5}}$. 
Squaring both sides: $\frac{b^4}{9-b^2} = \frac{16}{5} \implies 5b^4 = 16(9-b^2) \implies 5b^4 = 144 - 16b^2$. 
$5b^4 + 16b^2 - 144 = 0$. Let $y=b^2$. $5y^2+16y-144=0$. $y = \frac{-16 \pm \sqrt{16^2 - 4(5)(-144)}}{2(5)} = \frac{-16 \pm \sqrt{256 + 2880}}{10} = \frac{-16 \pm \sqrt{3136}}{10} = \frac{-16 \pm 56}{10}$. Since $y=b^2>0$, we take the positive root: $y = \frac{-16+56}{10} = \frac{40}{10} = 4$. So, $b^2 = 4$. 
The equation of the ellipse is $\frac{x^2}{9} + \frac{y^2}{4} = 1$. 
The equation of the tangent at a point $(x_1, y_1)$ is $\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1$. 
The point is $(\frac{3}{\sqrt{2}}, \frac{b}{\sqrt{2}}) = (\frac{3}{\sqrt{2}}, \frac{2}{\sqrt{2}})$ since $b=\sqrt{4}=2$. 
The tangent equation is $\frac{x(3/\sqrt{2})}{9} + \frac{y(2/\sqrt{2})}{4} = 1 \implies \frac{x}{3\sqrt{2}} + \frac{y}{2\sqrt{2}} = 1$. 
To find the slope, we rearrange this into $y=mx+c$ form. 
$\frac{y}{2\sqrt{2}} = 1 - \frac{x}{3\sqrt{2}} \implies y = 2\sqrt{2} - \frac{2\sqrt{2}}{3\sqrt{2}}x \implies y = -\frac{2}{3}x + 2\sqrt{2}$. 
The slope of the tangent is $-2/3$.