If xcos⁡α+ysin⁡α=4 is tangent to x225+y29=1 , then the value of α is

xcos⁡α+ysin⁡α=4∴ y=(−cot⁡α)x+4cosec⁡α∴ m=−cot⁡α,c=4cosec⁡α,a2=25,b2=9 Now c2=a2m2+b2∴ 16cosec2⁡α=25cot2⁡α+9∴ 161+cot2⁡α=25cot2⁡α+9∴ 7=9cot2⁡α⇒cot⁡α=73⇒tan⁡α=37∴ α=tan−1⁡(3/7)

If xcos⁡α+ysin⁡α=4 is tangent to x225+y29=1 , then the value of α is

xcos⁡α+ysin⁡α=4∴ y=(−cot⁡α)x+4cosec⁡α∴ m=−cot⁡α,c=4cosec⁡α,a2=25,b2=9 Now c2=a2m2+b2∴ 16cosec2⁡α=25cot2⁡α+9∴ 161+cot2⁡α=25cot2⁡α+9∴ 7=9cot2⁡α⇒cot⁡α=73⇒tan⁡α=37∴ α=tan−1⁡(3/7)

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$If x \cos α + y \sin α = 4 is tangent to \frac{x^{2}}{25} + \frac{y^{2}}{9} = 1, then the value of α is$

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$\tan^{- 1} (3 / \sqrt{7})$

$\tan^{- 1} (7 / 3)$

$\tan^{- 1} (\sqrt{3} / 7)$

$\tan^{- 1} (3 / 7)$

answer is A.

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Detailed Solution

$x \cos α + y \sin α = 4$

$\begin{array}{l} ∴ y = (- \cot α) x + 4 cosec α \\ ∴ m = - \cot α, c = 4 cosec α, a^{2} = 25, b^{2} = 9 \end{array}$

$Now c^{2} = a^{2} m^{2} + b^{2}$

$\begin{array}{l} ∴ 16 {cosec}^{2} α = 25 \cot^{2} α + 9 \\ ∴ 16 (1 + \cot^{2} α) = 25 \cot^{2} α + 9 \\ ∴ 7 = 9 \cot^{2} α \Rightarrow \cot α = \frac{\sqrt{7}}{3} \Rightarrow \tan α = \frac{3}{\sqrt{7}} \\ ∴ α = \tan^{- 1} (3 / \sqrt{7}) \end{array}$