Let one end of a focal chord of the parabola y² = 16x be (16, 16). If P(α, β) divides this focal chord internally in the ratio 5 : 2, then the minimum value of α + β is equal t

The four standard parabolas and their key properties:

Equation	Axis	Focus	Directrix	Parametric
$y^2=4ax$	x-axis	$(a,0)$	$x=-a$	$(at^2,2at)$
$y^2=-4ax$	x-axis	$(-a,0)$	$x=a$	$(-at^2,2at)$
$x^2=4ay$	y-axis	$(0,a)$	$y=-a$	$(2at,at^2)$
$x^2=-4ay$	y-axis	$(0,-a)$	$y=a$	$(2at,-at^2)$

For $y^2=16x$: $a=4$, focus=$(4,0)$, directrix $x=-4$, latus rectum length $=4a=16$.

A focal chord is any chord of a parabola that passes through the focus.

Parametric endpoints: If one end has parameter $t_1$, the other end has parameter $t_2$ where: This means $t_2 = -1/t_1$ — a very important result.

Other important focal chord results:
• Length of focal chord $= a\!\left(t_1 – t_2\right)^2 = a\!\left(t_1 + \dfrac{1}{t_1}\right)^2$
• Minimum length of focal chord = latus rectum $= 4a$ (when $t_1 = \pm 1$)
• The tangents at the endpoints of a focal chord are perpendicular (meet at directrix)
• Foot of directrix perpendiculars from endpoints: these perpendicular feet and the two endpoints form a rectangle

If point $P$ divides the line segment joining $A(x_1,y_1)$ and $B(x_2,y_2)$ internally in ratio $m:n$, then: $$P = \left(\frac{mx_2 + nx_1}{m+n},\; \frac{my_2 + ny_1}{m+n}\right)$$ Remember: In $m:n$, the point is closer to the side with larger number. Here $5:2$ means $P$ is closer to $B(1,-4)$ than to $A(16,16)$.

For external division (ratio $m:n$ externally): $$P = \left(\frac{mx_2 – nx_1}{m-n},\; \frac{my_2 – ny_1}{m-n}\right)$$

The question says “minimum value of $\alpha+\beta$” which might suggest $P$ can vary. But here the focal chord is uniquely determined (one end is fixed at $(16,16)$), and the ratio $5:2$ is also fixed. So there is only one possible $P$.

However, one could interpret the ratio as $5:2$ either from $(16,16)$ end or from the other end, giving two possible positions of $P$:
• $P_1$: divides $(16,16)$ to $(1,-4)$ in $5:2$ → $\alpha+\beta = 37/7 + 12/7 = 49/7 = 7$
• $P_2$: divides $(16,16)$ to $(1,-4)$ in $2:5$ → $\alpha = (2\cdot1+5\cdot16)/7 = 82/7$, $\beta=(2\cdot(-4)+5\cdot16)/7 = 72/7$, sum $= 154/7 = 22$

Minimum of $\{7, 22\}$ = 7 ✓. This explains why “minimum” is used in the question.

The latus rectum is the focal chord perpendicular to the axis of the parabola.

For $y^2 = 4ax$:
• Endpoints of latus rectum: $(a, 2a)$ and $(a, -2a)$
• Length $= 4a$
• For $y^2=16x$: endpoints are $(4,8)$ and $(4,-8)$, length $= 16$

The latus rectum is the shortest focal chord of any parabola. All other focal chords are longer. This is a standard JEE result.

The focal distance (distance from focus to a point on the parabola) for $y^2=4ax$ at point $(x_1,y_1)$: $$\text{Focal distance} = x_1 + a$$ This uses the reflective property: distance from point to focus = distance from point to directrix.

For $(16,16)$ on $y^2=16x$: focal distance $= 16 + 4 = 20$.
For $(1,-4)$ on $y^2=16x$: focal distance $= 1 + 4 = 5$.

Harmonic mean property: If $r_1$ and $r_2$ are focal distances of the two ends of a focal chord, then: $$\frac{1}{r_1} + \frac{1}{r_2} = \frac{1}{a} \quad \text{(semi-latus rectum)}$$ Check: $1/20 + 1/5 = 1/20 + 4/20 = 5/20 = 1/4 = 1/a$ ✓

❌ Mistake 1: Using $t_1 t_2 = 1$ instead of $t_1 t_2 = -1$ for focal chord. The negative sign is essential — without it the chord won’t pass through focus.

❌ Mistake 2: Taking $t_1 = -2$ (negative) for point $(16,16)$. Since $y = 16 > 0$, we need $8t_1 > 0$, so $t_1 = +2$.

❌ Mistake 3: In section formula, using ratio in wrong order. $5:2$ from A to B means: $\alpha = (5\cdot x_B + 2\cdot x_A)/7$, not $(5\cdot x_A + 2\cdot x_B)/7$.

❌ Mistake 4: Not checking both interpretations of $5:2$ ratio when question asks for “minimum”.

$y^2=4ax$: parametric $(at^2,2at)$ Focal chord: $t_1t_2=-1$ Internal division: $(mx_2+nx_1)/(m+n)$ Focal distance $= x+a$ $1/r_1+1/r_2=1/a$ LR length $=4a$