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Locus of the Points Equidistant From a Given Point

A straight li...

Question

A straight line through the vertex P of the $△ P Q R$ intersects the side QR at S and the circumcircle of the $△ P Q R$ at T. If S is not the centre of the circumcircle, then

\frac{1}{P S} + \frac{1}{S T} < \frac{2}{\sqrt{Q S \cdot S R}}

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\frac{1}{P S} + \frac{1}{S T} > \frac{2}{\sqrt{Q S \cdot S R}}

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\frac{1}{P S} + \frac{1}{S T} < \frac{4}{Q R}

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\frac{1}{P S} + \frac{1}{S T} > \frac{4}{Q R}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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Solution

The correct options are
B $\frac{1}{P S} + \frac{1}{S T} > \frac{2}{\sqrt{Q S \cdot S R}}$
D $\frac{1}{P S} + \frac{1}{S T} > \frac{4}{Q R}$
Since $S$ is not the centre of the circumcenter $P S \neq S T, Q S \neq S R$
Since $A . M . > G . M .$
$\frac{1}{2} (\frac{1}{P S} + \frac{1}{S T}) > \frac{1}{\sqrt{P S \times S T}} \Rightarrow \frac{1}{P S} + \frac{1}{S T} > \frac{2}{\sqrt{Q S \times S T}} [∵ P S \times S T = Q S \times S R]$
We know that if $x + y = k$ , a constant then $x y$ is maximum if $x = y = \frac{k}{2}$
Thus $\Rightarrow \frac{1}{Q S \times S R} > \frac{4}{{(Q S)}^{2}} \Rightarrow \frac{1}{P S} + \frac{1}{S T} > \frac{2}{\sqrt{Q S \times S R}} > \frac{4}{Q R}$

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Similar questions

{PR}^{2} = {PS}^{2} + QR \times ST + {(\frac{QR}{2})}^{2}

{PQ}^{2} = {PS}^{2} - QR \times ST + {(\frac{QR}{2})}^{2}

△ P Q R,

P S

P

Q R

Q R

S .

P S : Q S : R S = 2 : 4 : 1,

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P

P Q R

Q R

S

P Q R

T

S

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