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Let $\mathrm{\mathrm{P}: y^{2}=4 a x, a>0}$ be a parabola with focus $\mathrm{S}$ . Let the tangents to the parabola $\mathrm{P}$ make an angle of $\mathrm{\frac{\pi}{4}}$ with the line $\mathrm{y=3x+5}$ touch the parabola $\mathrm{P}$ at $\mathrm{A}$ and $\mathrm{B}$ . Then the value of $\mathrm{a}$ for which $\mathrm{A}$ , $\mathrm{B}$ and $\mathrm{S}$ are collinear is
Option: 1
$\mathrm{8\: only}$

Option: 2
$\mathrm{2\: only}$

Option: 3
$\mathrm{\frac{1}{4}\: only}$

Option: 4
$\mathrm{any\: a> 0}$

Answers (1)

best_answer

Tangents $\mathrm{T_{1}\: and\: T_{2}}$ make angle $\mathrm{\frac{\pi}{4}}$ with line $\mathrm{y=3x+5}$ means that tangents are perpendicular

Using property of parabola, point of contact of perpendicular tangents always form a focal chord. So $\mathrm{A,S\: and\: B}$ will always be collinear for any value of $\mathrm{a> 0}$ .

Hence the answer is option 4

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