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JEE Main 202117 MarchEvening Shift
Exam
JEE Main
Year
2021
Date
17 March
Shift
Evening Shift
Subject
Mathematics
Chapter
Ellipse

JEE Main 2021 · 17 March · Evening Shift · Mathematics · Ellipse

Let L be a tangent line to the parabola y ^ 2 = 4x - 20 at (6, 2) ....

Let LbeatangentlinetotheparabolaL be a tangent line to the parabola{y}^{2} = 4x - 20at(6,2).at \left(6, 2\right) . If LL is also a tangent to the ellipse x22+y2b=1,\dfrac{{x}^{2}}{2} + \dfrac{{y}^{2}}{b} = 1, then the value of bb is equal to :

Options

  1. A1111
  2. B1414
  3. C1616
  4. D2020

Correct answer

B. 1414

Step-by-step solution

The equation of a tangent to a parabola y2=4a(x−h){y}^{2} = 4a\left(x - h\right) at a point \left({x}_{1}, {y}_{1}\right)onitison it isy{y}_{1} = 4a\left(\left(\dfrac{x + {x}_{1}}{2}\right) - h\right) .</p><p>Hence,theequationoftangenttoparabola</p><p>Hence, the equation of tangent to parabola{y}^{2} = 4x - 20at(6,2)at \left(6, 2\right) is 2y=2(x+6)−202y = 2 ( x + 6 ) - 20

⇒y=x−4.⇒ y = x - 4 .

Also, a line y=mx+cy = mx + c is a tangent to the ellipse x2a2+y2b2=1\dfrac{{x}^{2}}{{a}^{2}} + \dfrac{{y}^{2}}{{b}^{2}} = 1 if c2=a2m2+b2.{c}^{2} = {a}^{2}{m}^{2} + {b}^{2} .

Hence, by using the condition of tangency for ellipsex22+y2b=1andtheline\dfrac{{x}^{2}}{2} + \dfrac{{y}^{2}}{b} = 1 and the liney = x - 4,weget</p><p>we get</p><p>16 = 2 ( 1{)}^{2} + b</p><p></p><p>⇒ b = 14 .$

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