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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Parabola

NTA Abhyas 2020 · Mathematics · Parabola

Let the tangents PQ and PR are drawn to y^ 2 = 4ax from any point P...

Let the tangents PQPQ and PRPR are drawn to y2=4axy^{2} = 4ax from any point PP on the line x+4a=0.x + 4a = 0. The angle subtended by the chord of contact QRQR at the vertex of the parabola y2=4axy^{2} = 4ax is

Options

  1. Aπ4\dfrac{\pi}{4}
  2. Bπ3\dfrac{\pi}{3}
  3. Cπ2\dfrac{\pi}{2}
  4. Dπ6\dfrac{\pi}{6}

Correct answer

C. π2\dfrac{\pi}{2}

Step-by-step solution

Let, Q=(at12,2at1)Q = \left(a{t}_{1}^{2}, 2a{t}_{1}\right) and R=(at22,2at2)R = \left(a{t}_{2}^{2}, 2a{t}_{2}\right), then P=(at1t2,a(t1+t2))P = \left(a{t}_{1}{t}_{2}, a\left({t}_{1} + {t}_{2}\right)\right) which lies on x+4a=0x + 4a = 0
⇒at1t2=−4a⇒t1t2=−4⇒ a{t}_{1}{t}_{2} = - 4a ⇒ {t}_{1}{t}_{2} = - 4
Now, the vertex of the parabola y2=4ax{y}^{2} = 4ax is the origin O(0,0)O\left(0,0\right)
⇒⇒slope of OQ=2t1OQ = \dfrac{2}{{t}_{1}} & slope of OR=2t2OR = \dfrac{2}{{t}_{2}}
⇒2t1×2t2=4t1t2=4−4=−1⇒ \dfrac{2}{{t}_{1}} \times \dfrac{2}{{t}_{2}} = \dfrac{4}{{t}_{1}{t}_{2}} = \dfrac{4}{- 4} = - 1
⇒OQ⇒ OQ is perpendicular to OROR
⇒∠QOR=90∘=π2⇒ ∠QOR = 9{0}^{∘} = \dfrac{\pi}{2}
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