Question

$\large \boxed{\sf \red{Question}}$
Find the angle subtended at the origin by the chord 4px - √2py = 4√2b of the curve px² - 4by = 0.​

Answer 1

Step-by-step explanation:

Find the angle subtended at the origin by the chord 4px - √2py = 4√2b of the curve px² - 4by = 0.

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Answer 2

$</p><p> \green{\large\underline{\sf{Solution-}}} </p><p>$

Given chord is

$rm :\longmapsto\:4px - \sqrt{2}py = 4 \sqrt{2}$

can be rewritten as

$\rm :\longmapsto\:\dfrac{4px - \sqrt{2}py }{4 \sqrt{2} b} = 1 - - - (1)$

Now, Given curve is

$rm :\longmapsto\: {px}^{2} - 4by = 0$

can be further rewritten as to make it homogeneous,

$\rm :\longmapsto\: {px}^{2} - 4by \times 1= 0$

On substituting the value from equation (1), we get

$\rm :\longmapsto\: {px}^{2} - 4by \times \dfrac{4px - \sqrt{2}py }{4 \sqrt{2} b}= 0$

$\rm :\longmapsto\: {px}^{2} - \dfrac{4pxy - \sqrt{2}p {y}^{2} }{\sqrt{2} }= 0$

$\rm :\longmapsto\: {px}^{2} - \dfrac{ \sqrt{2} \: (2 \sqrt{2} pxy - p {y}^{2})}{\sqrt{2} }= 0$

$\rm :\longmapsto\: {px}^{2} - 2 \sqrt{2} pxy + p {y}^{2}= 0$

$\rm :\longmapsto\: p({x}^{2} - 2 \sqrt{2} xy + {y}^{2})= 0$

$\rm :\longmapsto\: {x}^{2} - 2 \sqrt{2} xy + {y}^{2}= 0$

This represents the equation of pair of straight lines passes through the origin.

So,

On comparing with general equation

$\red{\rm :\longmapsto\: {ax}^{2} + 2hxy + {by}^{2} = 0}$

we get

$</p><p>\rm :\longmapsto\:a = 1$

$rm :\longmapsto\:h = - \sqrt{2}$

$\rm :\longmapsto\:b = 1$

So, angle between pair of lines is given by

$\rm :\longmapsto\:\boxed{\tt{ tan \theta = \frac{2 \sqrt{ {h}^{2} - ab} }{a + b} \: }}$

So, on substituting the values, we get

$\rm :\longmapsto\: tan \theta = \dfrac{2 \sqrt{ {( - \sqrt{2} )}^{2} - 1 \times 1} }{1 + 1} \:$

$\rm :\longmapsto\: tan \theta = \dfrac{2 \sqrt{ {2 - 1}} }{2} \:$

$\rm :\longmapsto\: tan \theta = 1:⟼tanθ=1</p><p></p><p>\bf\implies \: \theta = \dfrac{\pi}{4} ⟹θ= \pi4$