Parabola flash cards

1. 1.For a parabola $y^2=4ax$, give the coordinates of the focus and the equation of the directrix.

   Focus $=(a,0)$; Directrix: $x=-a$.

   Hint: Latus rectum length is $4a$.

2. 2.State the condition for the line $y=mx+c$ to be a tangent to the parabola $y^2=4ax$, and write the point of contact.

   Tangency condition: $c=\dfrac{a}{m}$; point of contact $=\left(\dfrac{a}{m^2},\dfrac{2a}{m}\right)$.

   Hint: Substitute the line into $y^2=4ax$ and set discriminant $=0$.

3. 3.For the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ $(a>b)$, write the eccentricity relation and the length of the latus rectum.

   $b^2=a^2(1-e^2)$; Latus rectum $=\dfrac{2b^2}{a}$.

   Hint: Foci are at $(\pm ae,0)$.

4. 4.State the defining property (sum/difference of distances) that characterizes an ellipse and a hyperbola with foci $S_1,S_2$.

   Ellipse: $SP+S'P=2a$ (constant sum); Hyperbola: $|SP-S'P|=2a$ (constant difference).

   Hint: Both are loci definitions in terms of distances to the two foci.

5. 5.For a hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$, write the relation between $a,b,e$ and the equations of its asymptotes.

   $b^2=a^2(e^2-1)$; Asymptotes: $y=\pm\dfrac{b}{a}x$.

   Hint: Eccentricity of a hyperbola is always $e>1$.

6. 6.For the parabola $y^2=4ax$, a focal chord meets the curve at parameters $t_1$ and $t_2$. Derive the relation between $t_1$ and $t_2$, and hence find $t_2$ if $t_1=2$.

   The chord joining $(at_1^2,2at_1)$ and $(at_2^2,2at_2)$ is $y(t_1+t_2)=2x+2at_1t_2$. It is a focal chord iff it passes through $(a,0)$: $0=2a+2at_1t_2\Rightarrow t_1t_2=-1$. For $t_1=2$, $t_2=-\dfrac12$.

   Hint: Substitute the focus $(a,0)$ into the parametric chord equation.

7. 7.A focal chord $PQ$ of $y^2=4ax$ has focal distance $SP=\dfrac{9a}{4}$. Find $SQ$.

   For any focal chord, $\dfrac1{SP}+\dfrac1{SQ}=\dfrac1a$ (the harmonic mean of the two focal segments equals the semi-latus rectum $2a$). This follows since $SP=a(1+t_1^2)$, $SQ=a(1+t_2^2)$ with $t_1t_2=-1$. So $\dfrac1{SQ}=\dfrac1a-\dfrac4{9a}=\dfrac5{9a}\Rightarrow SQ=\dfrac{9a}5$.

   Hint: Recall the harmonic mean of the two focal segments equals $2a$.

8. 8.Prove that the tangents at the extremities of a focal chord of $y^2=4ax$ intersect at right angles on the directrix.

   Tangents at $t_1,t_2$ meet at $(at_1t_2,\,a(t_1+t_2))$. For a focal chord $t_1t_2=-1$, so the point is $(-a,\,a(t_1+t_2))$, which lies on $x=-a$, the directrix. Their slopes are $1/t_1,1/t_2$ with product $\dfrac1{t_1t_2}=-1$, so the tangents are perpendicular. Result: perpendicular tangents meeting on the directrix.

   Hint: Use the tangent-intersection formula together with $t_1t_2=-1$.

9. 9.Find the minimum possible length of a focal chord of $y^2=4ax$, and identify which chord attains it.

   Length $=SP+SQ=a(2+t_1^2+t_2^2)$; with $t_2=-1/t_1$ this becomes $a\left(t_1+\dfrac1{t_1}\right)^2$, which is minimized when $|t_1|=1$, giving length $4a$. This minimum focal chord is exactly the latus rectum.

   Hint: Write the length purely in terms of $t_1$ using $t_1t_2=-1$, then minimize.

10. 10.Show that a focal chord of $y^2=4ax$ inclined at angle $\theta$ to the axis has length $4a\csc^2\theta$.

    The focal chord line is $y=\tan\theta\,(x-a)$. Substituting into $y^2=4ax$ gives $\tan^2\theta\,(x-a)^2=4ax$, a quadratic in $x$ whose roots satisfy $x_1+x_2=2a+\dfrac{4a}{\tan^2\theta}$. Since length $=x_1+x_2+2a=4a+4a\cot^2\theta=4a(1+\cot^2\theta)=4a\csc^2\theta$.

    Hint: Form the quadratic in $x$ from the line through the focus and sum its roots.

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