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JEE Main 202225 JuneMorning Shift
Exam
JEE Main
Year
2022
Date
25 June
Shift
Morning Shift
Subject
Mathematics
Chapter
Parabola

JEE Main 2022 · 25 June · Morning Shift · Mathematics · Parabola

If y = m _ 1 x + c _ 1 and y = m _ 2 x + c _ 2 , m _ 1 ≠ m _ 2 are...

If y=m1x+c1y = {m}_{1}x + {c}_{1} and y=m2x+c2,m1≠m2y = {m}_{2}x + {c}_{2}, {m}_{1} \ne {m}_{2} are two common tangents of circle x2+y2=2{x}^{2} + {y}^{2} = 2 and parabola y2=x{y}^{2} = x, then the value of 8∣m1m2∣8\left|{m}_{1}{m}_{2}\right| is equal toIf y = m _ 1 x + c _ 1 and y = m _ 2 x + c _ 2 , m _ 1 ≠ m _ 2 are two common tangents of circle x ^ 2 + y ^ 2 = 2 and

Options

  1. A32−43\sqrt{2} - 4
  2. B62−46\sqrt{2} - 4
  3. C−5+62- 5 + 6\sqrt{2}
  4. D3+423 + 4\sqrt{2}

Correct answer

A. 32−43\sqrt{2} - 4

Step-by-step solution

If y = m _ 1 x + c _ 1 and y = m _ 2 x + c _ 2 , m _ 1 ≠ m _ 2 are two common tangents of circle x ^ 2 + y ^ 2 = 2 and

Equation of the tangent to the parabola can be written as y=mx+(14m)y = mx + \left(\dfrac{1}{4m}\right) or y−mx−14m=0y - mx - \dfrac{1}{4m} = 0

Now, the perpendicular distance from \left(0, 0\right)tothetangentisequaltotheradiusofthegivencircle,soto the tangent is equal to the radius of the given circle, soOM = \sqrt{2} ⇒ \left|\dfrac{\dfrac{- 1}{4m}}{\sqrt{1 + {m}^{2}}}\right| = \sqrt{2}</p><p></p><p>⇒ 2\left(16{m}^{2}\right)\left(1 + {m}^{2}\right) = 1 ⇒ 32{m}^{4} + 32{m}^{2} - 1 = 0</p><p></p><p>⇒ {m}^{2} = \dfrac{- {2}^{5} + {2}^{3}\sqrt{16 + 2}}{2 \times {2}^{5}}ignoringthenegativesignasitissquarefunction.</p><p>Now∣8m1m2∣=∣8m2∣ignoring the negative sign as it is square function.</p><p>Now \left|8{m}_{1}{m}_{2}\right| = \left|8{m}^{2}\right|

= \left|\dfrac{- {2}^{5} + {2}^{3}\sqrt{16 + 2}}{{2}^{3}}\right| = 3\sqrt{2} - 4

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