# Least value of slope of tangent to hyperbola

While studying hyperbola, I came across a question:

Let $$y=mx+c$$ is a tangent to a hyperbola $$\cfrac{x^2}{ \lambda^2} -\cfrac{y^2}{( \lambda^3+ \lambda^2+\lambda)^2} = 1$$Find least value of $$16m^2$$.

My attempt:

As $$y =mx +c$$ is tangent so $$c^2=a^2m^2-b^2$$ then I put value of $$a$$ and $$b$$ and I take derivative of it but there is no information about $$\lambda$$. How should I proceed?

• Have you tried with a given hyperbola and some points? Jul 3, 2019 at 9:53
• Where does $c^2=a^2m^2-b^2$ come from? What do $a$ and $b$ stand for? Jul 7, 2019 at 3:24
• a=lamda ,b=lambda^3+lamda^2+lamda Jul 7, 2019 at 3:27
• Least value of 16m^2 given in answer is 9 Jul 7, 2019 at 3:31
• If you are looking for the least slope, then you can find the slope right under the right vertex; it will have a slope of $-\infty$. Are you instead looking for the least absolute value of the slope? Jul 7, 2019 at 3:38

As you have said $$c^2 = a^2 m^2 - b^2$$ ,On placing values$$c = \pm \sqrt{\lambda^2 m^2 - (\lambda + \lambda^2 + \lambda^3)^2}$$ $$\implies {\lambda^2 m^2 - (\lambda + \lambda^2 + \lambda^3)^2} \ge 0$$ $$\implies \lambda^2( m^2 - (1 + \lambda + \lambda^2)^2 )\ge 0$$ Assuming $$\lambda \ne 0$$, $$m^2 \ge ( 1 + \lambda + \lambda^2)^2$$ As nothing is specified about the nature of values of $$\lambda$$, it can be assumed that $$\lambda \in \mathbb R - \{0\}$$. Least value of the polynomial $$1 + \lambda + \lambda^2$$ is $$\dfrac{3}{4}$$ $$\implies m^2 \ge \dfrac{9}{16} \implies 16 m^2 \ge 9$$
• Let $f( x) = 1 + x + x^2$, on differentiating and equating the derivative to zero(for finding minima, as there is not maxima), $f'(x) = 2x + 1 = 0 \implies x = - 0.5$ , place this in the $f(x)$, you will get the minima Jul 7, 2019 at 4:14