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JEE Main 20205 SeptemberEvening Shift
Exam
JEE Main
Year
2020
Date
5 September
Shift
Evening Shift
Subject
Mathematics
Chapter
Hyperbola

JEE Main 2020 · 5 September · Evening Shift · Mathematics · Hyperbola

If the line y = mx + c is a common tangent to the hyperbola x ^ 2...

If the line y=mx+cy = mx + c is a common tangent to the hyperbola x2100−y264=1\dfrac{{x}^{2}}{100} - \dfrac{{y}^{2}}{64} = 1 and the circle x2+y2=36,{x}^{2} + {y}^{2} = 36, then which one of the following is true?

Options

  1. Ac2=369{c}^{2} = 369
  2. B5m=45m = 4
  3. C4c2=3694{c}^{2} = 369
  4. D8m+5=08m + 5 = 0

Correct answer

C. 4c2=3694{c}^{2} = 369

Step-by-step solution

A line y=mx+cy = mx + c is a tangent to the hyperbola 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} and the line is tangent to the circle x2+y2=r2{x}^{2} + {y}^{2} = {r}^{2} is c2=r2(1+m2).{c}^{2} = {r}^{2}\left(1 + {m}^{2}\right) .

Hence, for the given line and the hyperbola we have, c2=100m2−64...(i){c}^{2} = 100{m}^{2} - 64 . . . \left(i\right) and for the line and the circle, we have

c2=36(1+m2)...(ii){c}^{2} = 36\left(1 + {m}^{2}\right) . . . \left(ii\right)

On comparing the above two equations, we get

100m2−64=36+36m2100{m}^{2} - 64 = 36 + 36{m}^{2}

64m2=10064{m}^{2} = 100

m2=10064{m}^{2} = \dfrac{100}{64} ⇒m=108=54.⇒ m = \dfrac{10}{8} = \dfrac{5}{4} .

⇒c2=36(1+2516)=3694.⇒ {c}^{2} = 36\left(1 + \dfrac{25}{16}\right) = \dfrac{369}{4} .

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