Parabola — Complete JEE Revision Capsule
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1. Definition & Standard Forms
Parabola — locus of a point equidistant from a fixed point (focus) and fixed line (directrix).
Standard forms (vertex at origin):
- Right opening: $\,y^{2}=4ax\,$ (focus $(a,0)$, directrix $x=-a$)
- Left opening: $\,y^{2}=-4ax\,$
- Upward: $\,x^{2}=4ay\,$ (focus $(0,a)$)
- Downward: $\,x^{2}=-4ay\,$
Vertex form (shifted): $(y-k)^2 = 4a(x-h)$ → vertex $(h,k)$, focus $(h+a,k)$.
2. Quick Geometry & Important Quantities
- Focus: $(a,0)$ for $y^2=4ax$.
- Directrix: $x=-a$ for $y^2=4ax$.
- Axis: the line through vertex and focus (here $x$-axis).
- Latus rectum: length $=4a$. Endpoints $(a,2a)$ and $(a,-2a)$ for $y^2=4ax$.
- Eccentricity: $e=1$ (parabola).
- Reflective property: ray from focus reflects to become parallel to axis.
3. Parametric Form (Powerful for JEE)
For $y^{2}=4ax$ use parameter $t$:
$$ x = at^{2},\qquad y = 2at. $$
Useful relations:
- Slope at parameter $t$: $$\displaystyle \frac{dy}{dx}=\frac{1}{t}.$$
- Point corresponding to $t$ has coordinates $(at^{2},2at)$.
- Eliminate $t$: $t=\dfrac{y}{2a}$ and $x=a\left(\dfrac{y}{2a}\right)^2$ (consistency check).
4. Tangents — three standard forms
Tangent at parameter $t$ (parametric form):
$$ \boxed{\,ty = x + at^{2}\,} $$
Slope form (with slope $m$): since $m=\dfrac{1}{t}$,
$$ \boxed{\,y = mx + \dfrac{a}{m}\,} $$
Point form (through point $(x_1,y_1)$ on parabola):
If $(x_1,y_1)$ lies on $y^2=4ax$, tangent is $$ \boxed{\,yy_1 = a(x + x_1)\,}. $$
Check for tangency: Substitute line $y=mx+c$ into parabola and require discriminant $=0$. That gives $c=\dfrac{a}{m}$ for tangency.
5. Normals
Normal at parameter $t$ (standard):
$$ \boxed{\,y = -tx + 2at + at^{3}\,} $$
Remarks: Equation of normal leads to cubic in $t$ when you ask for normals through a fixed point — typical JEE/theme problem. The three roots correspond to three normals through a point (real/complex depending on point).
6. Chords & Focal Chords
Chord joining points with parameters $t_1,t_2$: the general chord equation (obtain via two-point form) can be written as
$$ (t_1+t_2)\,y = 2x + a(t_1+t_2)t_1 t_2 \quad\text{(use algebraic elimination via parametric coords)}. $$
Focal chord: a chord through focus has parameters $t$ and $-\dfrac{1}{t}$. (Important property: product $t_1t_2=-1$.)
Latus rectum is the focal chord with $t=\pm1$; length $=4a$.
7. Important Lengths & Distances
- Distance of $(x,y)$ from focus $(a,0)$: $\sqrt{(x-a)^2 + y^2}$.
- Distance from directrix $x=-a$: $|x + a|$.
- Point $(x,y)$ lies on parabola iff distances equal: $\sqrt{(x-a)^2+y^2}=|x+a|$.
- Length of tangent from point $(x_1,y_1)$ (if external): find tangent line and foot; use algebra — preferred: use condition of tangency or power method.
8. Useful Identities & Quick Tricks
- Product of slopes of two perpendicular tangents: not applicable directly (parabola has no pair of perpendicular tangents everywhere).
- Midpoint $(X,Y)$ of chord joining $t_1,t_2$ satisfies relation derived from parameters; when $t_1+t_2=0$ chord is focal chord's symmetric case.
- To find tangent with given slope $m$: use $y=mx+\dfrac{a}{m}$ instantly.
- To test if line $y=mx+c$ is tangent: check $c=\dfrac{a}{m}$ (for $y^2=4ax$).
9. Common JEE Pain Points & How to Beat Them
- Mistaking tangent & chord forms: memorize the three tangent templates (parametric, slope, point forms) and check consistency with point/parameter.
- Sign errors with left/up/down opening: switch $a\to -a$ or change variable roles (swap x/y) — always convert to canonical form first.
- Normals produce cubic equations: don’t panic — treat cubic in $t$ and use symmetry/tricks (e.g., if normal passes through vertex or focus, special $t$ values appear).
- Parametric traps: when using $t$ keep it consistent: $x=at^2$, $y=2at$.
- Area/perimeter questions: use parametric integration if needed; param helps compute arc lengths and sector areas elegantly for many JEE problems.
10. Short Worked Examples (High-yield)
Ex 1 — Tangent with given slope: Find equation of tangent to $y^2=4ax$ with slope $m$.
Solution: Use slope form $$y=mx+\dfrac{a}{m}.$$
Ex 2 — Tangent at point: Find tangent at $(at_1^2,2at_1)$.
Solution: Parametric tangent: $$t_1y = x + at_1^2.$$
Ex 3 — Normal through a given point: Find normals to parabola passing through $(h,k)$. (Sketch)
Method: Substitute parametric normal $$y = -tx + 2at + at^3$$ and equate to pass through $(h,k)$ to get cubic in $t$: $$ k = -th + 2at + at^3. $$ Solve cubic (analytically or use symmetry/tricks for special points).
11. Practice Problems (Try first)
- Find equation(s) of tangent(s) parallel to line $y=3x+5$ for $y^2=4ax$.
- Find equation of normal to $y^2=4ax$ at $(4,4)$ (assume appropriate $a$) and check whether it passes through a given point.
- Show that the chord joining $(at_1^2,2at_1)$ and $(at_2^2,2at_2)$ has equation $(t_1+t_2)y = 2x + a t_1 t_2 (t_1+t_2)$ and verify for special case $t_1=-t_2$.
- Find locus of midpoints of chords of parabola which are perpendicular to axis.
- Length of latus rectum and coordinates of its endpoints for $y^2=12x$.
Show Answers / Hints
- Tangents parallel to $y=3x+5$ have slope $m=3$ → use $y=3x + a/3$ and pick $a$ accordingly if given.
- Compute $a$ from point lying on parabola then use normal formula to get cubic and check point substitution.
- Use parametric coords and two-point form to derive chord eqn; for $t_1=-t_2$ it reduces to $y=0$ (axis) as expected.
- Midpoint of horizontal chord has fixed y=0 (for chords perpendicular to axis) — derive parametrically.
- For $y^2=12x$, $4a=12\Rightarrow a=3$, latus rectum length $=4a=12$, endpoints $(3,6)$ and $(3,-6)$.
12. One-line Takeaways
- Memorize parametric form $x=at^2,\ y=2at$ — it unlocks tangents, normals, chords fast.
- Tangent templates: $ty=x+at^2$, $y=mx+\dfrac{a}{m}$, $yy_1=a(x+x_1)$ — use the one that fits.
- Normals → cubic in $t$; use symmetry and special values to simplify.
- Check signs & orientation (left/right/up/down) before applying formulae.
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