Hyperbola flash cards

1. 1.For the hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$, what is the relation between $a$, $b$, and eccentricity $e$, and what are the foci/directrices?

   $b^2=a^2(e^2-1)$, so $e=\sqrt{1+\dfrac{b^2}{a^2}}>1$. Foci: $(\pm ae,0)$. Directrices: $x=\pm\dfrac{a}{e}$.

   Hint: Opposite sign convention vs ellipse: $b^2=a^2(e^2-1)$

2. 2.What is the condition on $c$ for the line $y=mx+c$ to be a tangent to $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$, and what is the point of contact?

   Tangent condition: $c^2=a^2m^2-b^2$. Point of contact: $\left(\dfrac{-a^2m}{c},\dfrac{-b^2}{c}\right)$. Tangent line: $y=mx\pm\sqrt{a^2m^2-b^2}$.

   Hint: Compare with ellipse's $c^2=a^2m^2+b^2$

3. 3.What are the equations of the asymptotes of $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$, and how is the conjugate hyperbola related to eccentricities $e_1,e_2$?

   Asymptotes: $\dfrac{x}{a}\pm\dfrac{y}{b}=0$, i.e. $y=\pm\dfrac{b}{a}x$. Conjugate hyperbola: $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=-1$. Relation: $\dfrac{1}{e_1^2}+\dfrac{1}{e_2^2}=1$.

   Hint: Conjugate flips the RHS sign

4. 4.For a rectangular (equilateral) hyperbola $xy=c^2$, what is the equation of the tangent at parametric point $\left(ct,\dfrac{c}{t}\right)$, and what is its eccentricity?

   Tangent: $\dfrac{x}{t}+yt=2c$. Eccentricity $e=\sqrt{2}$ (asymptotes are the coordinate axes, perpendicular to each other).

   Hint: Parametric form $(ct, c/t)$

5. 5.For a point $P$ on the hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ with foci $S,S'$, what is $|SP-S'P|$, and what is the length of the semi-latus rectum?

   $|SP-S'P|=2a$ (constant, defining property). Semi-latus rectum $=\dfrac{b^2}{a}$, so full latus rectum $=\dfrac{2b^2}{a}$.

   Hint: Difference of focal distances is constant (unlike ellipse's sum)

6. 6.Derive the equation of the tangent to the rectangular hyperbola $xy=c^2$ at the point $\left(ct,\dfrac{c}{t}\right)$.

   Differentiating $xy=c^2$ implicitly, $y+xy'=0 \Rightarrow y'=-\dfrac{y}{x}$, so at $\left(ct,\frac{c}{t}\right)$ the slope is $-\dfrac{1}{t^2}$. The tangent is $y-\dfrac{c}{t}=-\dfrac{1}{t^2}(x-ct)$, which simplifies to $x+t^2y=2ct$.

   Hint: Use implicit differentiation on $xy=c^2$ to get the slope first.

7. 7.Find the equation of the normal to $xy=c^2$ at the point $\left(ct,\dfrac{c}{t}\right)$.

   The tangent slope at $\left(ct,\frac{c}{t}\right)$ is $-\dfrac{1}{t^2}$, so the normal slope is $t^2$. The normal is $y-\dfrac{c}{t}=t^2(x-ct)$; multiplying through by $t$ and simplifying gives $xt^3-yt=ct^4-c$.

   Hint: Normal slope is the negative reciprocal of the tangent slope $-1/t^2$.

8. 8.Show that the eccentricity of any rectangular hyperbola is $\sqrt2$.

   A rectangular hyperbola has perpendicular asymptotes, which forces the semi-transverse and semi-conjugate axes to be equal, $a=b$. Then $e^2=1+\dfrac{b^2}{a^2}=1+1=2$, so $e=\sqrt2$.

   Hint: Perpendicular asymptotes force $a=b$; use $e^2=1+b^2/a^2$.

9. 9.Find the slope of the chord joining the points $\left(ct_1,\dfrac{c}{t_1}\right)$ and $\left(ct_2,\dfrac{c}{t_2}\right)$ on $xy=c^2$.

   Slope $=\dfrac{c/t_1-c/t_2}{ct_1-ct_2}=\dfrac{c(t_2-t_1)/(t_1t_2)}{c(t_1-t_2)}=\dfrac{-1}{t_1t_2}$. So the chord's slope is $-\dfrac{1}{t_1t_2}$, matching the tangent-slope formula as $t_2\to t_1$.

   Hint: Just compute rise over run and simplify using a common denominator.

10. 10.Write the equation of the chord joining $\left(ct_1,\dfrac{c}{t_1}\right)$ and $\left(ct_2,\dfrac{c}{t_2}\right)$ on $xy=c^2$.

    Using point-slope form with slope $-\dfrac{1}{t_1t_2}$ through $\left(ct_1,\frac{c}{t_1}\right)$: $y-\dfrac{c}{t_1}=-\dfrac{1}{t_1t_2}(x-ct_1)$. Clearing denominators gives $x+t_1t_2\,y=c(t_1+t_2)$, which correctly reduces to the tangent when $t_1=t_2=t$.

    Hint: Start from the chord slope $-1/(t_1t_2)$ found separately and use point-slope form.

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