The distance between the two points A and A’ which lie on y = 2 such that both the line segments AB and A’B (where B is the point (2, 3)) subtend angle&nbsp;<math><mrow><mfrac><mrow><mi>π</mi></mrow><mrow><mn>4</mn></mrow></mfrac></mrow></math>&nbsp;at the origin, is equal to:

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Maths Class 12th Straight Lines Maths NCERT Exemplar Solutions Class 12th Chapter One

The distance between the two points A and A’ which lie on y = 2 such that both the line segments AB and A’B (where B is the point (2, 3)) subtend angle $\frac{π}{4}$ at the origin, is equal to:

Option 1 -

Option 2 -

$\frac{48}{5}$

Option 3 -

$\frac{52}{5}$

Option 4 -

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Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 3

Detailed Solution:

Let A, A’ be (, 2) AB and A’B subtends $\frac{π}{4}$ angle at (0, 0) slope of OA = $\frac{2}{α}$

slope of OB = $\frac{3}{2}$

$t a n \frac{π}{4} = | \frac{\frac{2}{α} - \frac{3}{2}}{1 + \frac{2}{α} \cdot \frac{3}{2}} |$

$\Rightarrow 1 + \frac{3}{α} = \pm (\frac{4 - 3 α}{2 α})$

$\Rightarrow \frac{α + 3}{α} = \pm (\frac{4 - 3 α}{2 α}) \Rightarrow α = 10, - \frac{2}{5}$

now distance between A’A, (10, 2) & $(\frac{- 2}{5}, 2) i s \frac{52}{5}$

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Let point P : (h, k)

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Let equation of normal to x² = y at Q (t, t²) is x + 2ty = t + 2t³

It passes through the point (1, -1) so, 2t³ + 3t – 1 = 0

Let f(t) = 2t³ + 3t – 1 f $(\frac{1}{4}) f (\frac{1}{3}) < 0 \Rightarrow t \in (\frac{1}{4}, \frac{1}{3})$

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$t a n ? θ = \frac{10}{x} = \frac{h}{x_{2}} \Rightarrow x_{2} = \frac{h x}{10}$

$t a n ? ? = \frac{15}{x} = \frac{h}{x} \Rightarrow x_{1} = \frac{h x}{15}$

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The distance between the two points A and A’ which lie on y = 2 such that both the line segments AB and A’B (where B is the point (2, 3)) subtend angle $\frac{π}{4}$ at the origin, is equal to: