KVPY SX SB Exam 2016 Question Paper

Question : 7 of 120

Marks: +1, -0

Suppose the parabola

(y - k)^2 = 4 (x - h)

, with vertex A, passes through O = (0, 0) and L = (0, 2). Let D be an end point of the latus rectum. Let the y-axis intersect the axis of the parabola at P. Then ∠PDA is equal to

$\tan^{-1} 1/19$
$\tan^{-1} 2/19$
$\tan^{-1} 4/19$
$\tan^{-1} 8/19$

Validate

Solution:

Curve,

S : (y- k)^2 = 4(x - h)

LLR = 4 ; Clearly k = 1 ; ⇒ A(h, 1) & 'M' is focus (h + 1, 1)
So D (h +1, 3)

S_(0, 0) = 0 ⇒ k^2 = - 4h

⇒h=-1/4⇒D(3/4,3)

Now;

\tan α=|{m_1-m_2}/{1+m_1m_2}|

=|{8/3-2}/{1+8/3×2}|=2/19

where,

m_1={3-1}/{3/4-0}=8/3

m_2={3-1}/1=2

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