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Let one end of a focal chord of the parabola y 2 16 x b... | null (Mathematics) | DivineJEE

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Mathematics

JEE Mains 2026 21st January Shift 2

Mathematics

Let one end of a focal chord of the parabola

y^{2}=16 x

be (16, 16). If

\mathrm{P}(\alpha, \beta)

divides this focal chord internally in the ratio

5: 2

, then the minimum value of

\alpha+\beta

is equal to :

A

22

B

7

C

5

D

16

Correct Answer:

B

Explanation

\mathrm{y}^{2}=16 \mathrm{x}\\

Question Image

\\

∵ parameter of point A is

\mathrm{t}=2\\

⇒ Parameter of point

B

is

t=-\frac{1}{2}\\

⇒ Coordinates of B is

(1,-4)\\

Case 1:

Question Image

\\ \alpha=\frac{5+32}{7}=\frac{37}{7}\\

\beta=\frac{-20+32}{7}=\frac{12}{7}\\

\Rightarrow \alpha+\beta=7\\

Case 2 : A(16, 16)

Question Image

\\ \alpha=\frac{2+80}{7}

,

\\ \beta=\frac{-8+80}{7}\\

\alpha+\beta=22\\

So minimum value of

\alpha+\beta=7

Select an option to instantly check whether it is correct or wrong.

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