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

NTA Abhyas 2020 · Mathematics · Parabola

Let the normals at points A (4a, - 4a) and B (9a, - 6a) on the...

Let the normals at points A(4a,−4a)A \left(4a, - 4a\right) and B(9a,−6a)B \left(9a, - 6a\right) on the parabola y2=4axy^{2} = 4ax meet at the point P.P. The equation of the normal from PP on y2=4axy^{2} = 4ax (other than PAPA and PBPB)& is

Options

  1. A5x+y−135a=05x + y - 135a = 0
  2. B5x−y+115a=05x - y + 115a = 0
  3. C5x+y+115=05x + y + 115 = 0
  4. D5x−y−115a=05x - y - 115a = 0

Correct answer

A. 5x+y−135a=05x + y - 135a = 0

Step-by-step solution

Let, A = (at12,2at1)\left(a{t}_{1}^{2}, 2a{t}_{1}\right), B = \left(a{t}_{2}^{2}, 2a{t}_{2}\right)<br/>andthefootoftherequirednormalis(at32,2at3)<br />and the foot of the required normal is \left(a{t}_{3}^{2}, 2a{t}_{3}\right), then
2at1=−4a,2at2=−6aandt1+t2+t3=02a{t}_{1} = - 4a, 2a{t}_{2} = - 6a and {t}_{1} + {t}_{2} + {t}_{3} = 0
⇒t1=−2,t2=−3and−2−3+t3=0⇒ {t}_{1} = - 2, {t}_{2} = - 3 and - 2 - 3 + {t}_{3} = 0
⇒t3=5⇒ {t}_{3} = 5
Hence, the equation of the required normal is y=−t3x+2at3+at32y = - {t}_{3}x + 2a{t}_{3} + a{t}_{3}^{2}
⇒y=−5x+10a+125a⇒ y = - 5x + 10a + 125a
⇒y+5x−135a=0⇒ y + 5x - 135a = 0
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