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10-2. Parabola, Ellipse, Hyperbola
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If $2 x-y+1=0$ is a tangent to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{16}=1$, then which of the following $CANNOT$ be sides of a right angled triangle?
$[A]$ $2 a, 4,1$ $[B]$ $2 a, 8,1$ $[C]$ $a, 4,1$ $[D]$ $a, 4,2$
A
$A,D$
B
$B,D$
C
$B,C$
D
$B,C,D$
(IIT-2017)
Solution
Tangent to $\frac{x^2}{a^2}-\frac{y^2}{4^2}=1$ is
$\mathrm{y}=\mathrm{mx} \pm \sqrt{\mathrm{a}^2 \mathrm{~m}^2-16}$
Comparing with $\mathrm{y}=2 \mathrm{x}+1$
$\mathrm{m}=2$
$\Rightarrow 4 \mathrm{a}^2-16=1$
$\mathrm{a}^2=\frac{17}{4}$
$\mathrm{a}=\frac{\sqrt{17}}{2}$
Only $2 \mathrm{a}, 4,1$ are sides of a right-angled triangle
Standard 11
Mathematics
