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

Question

Let one end of a focal chord of the parabola $y^{2} = 16 x$ 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 to : [2026]

Smart Achievers · Accepted Answer

(1)

$y^{2} = 16 x$

$∵$ $parameter of point A is t = 2$

$\Rightarrow Parameter of point B is t = - \frac{1}{2}$

$\Rightarrow Coordinates of B is (1, - 4)$

$Case 1:$

$α = \frac{5 + 32}{7} = \frac{37}{7}$

$β = \frac{- 20 + 32}{7} = \frac{12}{7}$

$\Rightarrow α + β = 7$

$Case 2:$

$α = \frac{2 + 80}{7}, β = \frac{- 8 + 80}{7}$

$α + β = 22$

$So minimum value of α + β = 7$

Solution

Q.
Let one end of a focal chord of the parabola $y^{2} = 16 x$ 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 to : [2026]

1 7

2 5

3 16

4 22

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Solution

Q. 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 to : [2026]

1 7

2 5

3 16

4 22

Ans. (1) y2=16x ∵ parameter of point A is t=2 ⇒Parameter of point B is t=-12 ⇒Coordinates of B is (1,-4) Case 1: α=5+327=377 β=-20+327=127 ⇒α+β=7 Case 2: α=2+807, β=-8+807 α+β=22 So minimum value of α+β=7

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Q.
Let one end of a focal chord of the parabola $y^{2} = 16 x$ 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 to : [2026]