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10-2. Parabola, Ellipse, Hyperbola
normal
The equation of the tangents to the hyperbola $4x^2 -y^2 = 12$ are $y = 4x+ c_1 \,$$ \& \, y = 4x + c_2,$ then $|c_1 -c_2|$ is equal to -
A
$1$
B
$4$
C
$6$
D
$12$
Solution
$\left.\frac{d y}{d x}\right|_{(x, y)}=\frac{4 x_{1}}{y_{1}}=4$
$\Rightarrow y_{1}=x_{1}$
$\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$ lies on $4 \mathrm{x}^{2}-\mathrm{y}^{2}=12$
$\Rightarrow \mathrm{x}_{1}=\mathrm{y}_{1}=\pm 2$
$\therefore $ Equation of tangent lines with slope $4$ are
$y=4 x+6 $ and $ y=4 x-6$
$\Rightarrow\left|\mathrm{c}_{1}-\mathrm{c}_{2}\right|=12$
Standard 11
Mathematics
