Question

The value of m for which $y=mx+6$ is a tangent to the hyperbola $\frac{x^2}{100}-\frac{y^2}{49}=1$

• A. $\sqrt{\frac{20}{30}}$
• B. $\sqrt{\frac{20}{17}}$
• C. $\sqrt{\frac{17}{20}}$
• D. $\sqrt{\frac{3}{30}}$

Answer 1

Answer: (C)

$y=mx+6$ is a tangent to the hpyerbola
$\frac{x^2}{100}-\frac{y^2}{49}=1$ ...(i)
Tangent at $(x_1,y_1)$ to the hyperbola is given
by $\frac{x{x}_1}{100}-\frac{y{y}_1}{49}=1$ ...(ii)
Since equation $y=mx+6$ and (ii) represent same line, these equations should be identical and hence
$\frac{\frac{x_1}{100}}{-m}=\frac{\frac{-y_1}{49}}{1}=\frac{1}{6}\Rightarrow x_1=-\frac{100m}{6},y_1=-\frac{49}{6}$
$(x_1,y_1)$ lies on the hyperbola, therefore satisfies its equation
$\frac{{\left(-\frac{100m}{6}\right)}^2}{100}-\frac{{\left(-\frac{49}{6}\right)}^2}{49}=1\Rightarrow m=\sqrt{\frac{17}{20}}$