Number of correct statements among the following is . Then is ________ Statement I: Chord joining the points and of the parabola passes through focus of the parabola, length of the focal chord Statement II: Let and be two sets. Then Statement I

Question

Number of correct statements among the following is $k$ . Then $2 k$ is ________

Statement I: Chord joining the points $P (a t_{1}^{2}, 2 a t_{1})$ and $Q (a t_{2}^{2}, 2 a t_{2})$ of the parabola $y^{2} = 4 a x$ passes through focus of the parabola, length of the focal chord $P Q =$ $| a |$ ${(t_{1} + \frac{1}{t_{1}})}^{2} \geq 4 a$

Statement II: Let $P = {θ : \sin θ - \cos θ = \sqrt{2} \cos θ}$ and $Q = {θ : \sin θ + \cos θ = \sqrt{2} \sin θ}$ be two sets. Then $P = Q$

Statement III: If $a, b, c, d$ and $p$ are distinct non-zero real numbers such that $(a^{2} + b^{2} + c^{2}) p^{2} - 2 (a b + b c + c d) p + (b^{2} + c^{2} + d^{2}) \leq 0$ then $a, b, c, d$ are in G.P.

Smart Achievers · Accepted Answer

(6)

$Statement-I: Refer to chord properties of conics$

$Statement-II: P = {θ : \sin θ - \cos θ = \sqrt{2} \cos θ} \Rightarrow \sin θ = (\sqrt{2} + 1) \cos θ \Rightarrow \tan θ = \sqrt{2} + 1$

$Now, Q = {θ : \sin θ + \cos θ = \sqrt{2} \sin θ} \Rightarrow \cot θ = \sqrt{2} - 1 \Rightarrow \tan θ = \frac{1}{\sqrt{2} - 1} \times \frac{\sqrt{2} + 1}{\sqrt{2} + 1}$

$\Rightarrow \tan θ = \sqrt{2} + 1$

$∴$ $P = Q$

$Statement-III: {(a - b)}^{2} + {(b - c)}^{2} + {(c - b)}^{2} = 0$

Solution

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Solution

Ans. (6) Statement-I: Refer to chord properties of conics Statement-II: P={θ:sinθ-cosθ=2cosθ}⇒sinθ=(2+1)cosθ⇒tanθ=2+1 Now, Q={θ:sinθ+cosθ=2sinθ}⇒cotθ=2-1⇒tanθ=12-1×2+12+1 ⇒tanθ=2+1 ∴ P=Q Statement-III: (a-b)2+(b-c)2+(c-b)2=0

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