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JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola

જો વક્રો \(\frac{x^{2}}{16}+\frac{y^{2}}{9}=1\) અને \(x^{2}+y^{2}=12\) ના સામાન્ય સ્પર્શકની ઢાળ \(m\) હોય, તો \(12\,m^{2}=\dots\dots\dots\)

  1. A \(6\)
  2. B \(9\)
  3. C \(10\)
  4. D \(12\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(9\)

Step-by-step Solution

Detailed explanation

\(\frac{x^{2}}{16}+\frac{y^{2}}{9}=1\) equation of tangent to the ellipse is \(y=m x \pm \sqrt{a^{2} m^{2}+b^{2}}\) \(y=m x \pm \sqrt{16\; m^{2}+9}\) \(x^{2}+y^{2}=12\) equation of tangent to the circle is \(y=m x \pm \sqrt{12} \sqrt{1+m^{2}}\) for common tangent equate eq.…
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