If the tangent at the point (t2, 2t) on the parabola y2 = 4x is same as the normal drawn at point (5cos⁡θ,2sin⁡θ)on the ellipse 4x2 + 5y2 = 20, then

The equation of tangent is x – yt + t2 = 0→(1)The equation of normal at (5cos⁡θ,2sin⁡θ) on 4x2 +5y2 = 20 is(5sec⁡θ)x−(2cosec⁡θ)y=5−4⇒(5sec⁡θ)x−(2cosec⁡θ)y−1=0→(2)(1) and (2) represents same line then 5sec⁡θ1=−2cosec⁡θ−t=−1t2⇒t=25cot⁡θ and t=−12sin⁡θ⇒25cot⁡θ =−12sin⁡θ ⇒cosθ=-15 sub cosθ in t we get t=-15Hence θ=cos−1⁡−15 and t=−15

If the tangent at the point (t2, 2t) on the parabola y2 = 4x is same as the normal drawn at point (5cos⁡θ,2sin⁡θ)on the ellipse 4x2 + 5y2 = 20, then

The equation of tangent is x – yt + t2 = 0→(1)The equation of normal at (5cos⁡θ,2sin⁡θ) on 4x2 +5y2 = 20 is(5sec⁡θ)x−(2cosec⁡θ)y=5−4⇒(5sec⁡θ)x−(2cosec⁡θ)y−1=0→(2)(1) and (2) represents same line then 5sec⁡θ1=−2cosec⁡θ−t=−1t2⇒t=25cot⁡θ and t=−12sin⁡θ⇒25cot⁡θ =−12sin⁡θ ⇒cosθ=-15 sub cosθ in t we get t=-15Hence θ=cos−1⁡−15 and t=−15

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If the tangent at the point (t², 2t) on the parabola y² = 4x is same as the normal drawn at point $(\sqrt{5} \cos θ, 2 \sin θ)$ on the ellipse 4x² + 5y² = 20, then

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$θ = \cos^{- 1} (\frac{- 1}{\sqrt{5}})$

$θ = \cos^{- 1} (\frac{1}{\sqrt{5}})$

$t = \frac{- 2}{\sqrt{5}}$

$t = \frac{- 1}{\sqrt{5}}$

answer is A, D.

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Detailed Solution

The equation of tangent is x – yt + t² = 0 $\to$ (1)
The equation of normal at $(\sqrt{5} \cos θ, 2 \sin θ)$ on 4x² +5y² = 20 is
$\begin{aligned} (\sqrt{5} \sec θ) x - (2 cosec θ) y = 5 - 4 \\ \Rightarrow (\sqrt{5} \sec θ) x - (2 cosec θ) y - 1 = 0 \to (2) \end{aligned}$
(1) and (2) represents same line then
$\frac{\sqrt{5} \sec θ}{1} = \frac{- 2 cosec θ}{- t} = \frac{- 1}{t^{2}}$
$\Rightarrow t = \frac{2}{\sqrt{5}} \cot θ and t = \frac{- 1}{2} \sin θ$

$\Rightarrow \frac{2}{\sqrt{5}} \cot θ = \frac{- 1}{2} \sin θ \Rightarrow cosθ = \frac{- 1}{\sqrt{5}} sub cosθ in t we get t = \frac{- 1}{\sqrt{5}}$
Hence $θ = \cos^{- 1} (\frac{- 1}{\sqrt{5}}) and t = \frac{- 1}{\sqrt{5}}$