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10-2. Parabola, Ellipse, Hyperbola
hard
જો વક્રો $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$ અને $x^{2}+y^{2}=12$ ના સામાન્ય સ્પર્શકની ઢાળ $m$ હોય, તો $12\,m^{2}=\dots\dots\dots$
A
$6$
B
$9$
C
$10$
D
$12$
(JEE MAIN-2022)
Solution
$\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$
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. $(i)$ and $(ii)$
$\Rightarrow 16 \;m ^{2}+9=12\left(1+ m ^{2}\right)$
$16\; m^{2}-12\; m^{2}=3$
$4 \;m ^{2}=3$
$12 \;m ^{2}=9$
Standard 11
Mathematics
