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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?

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  • $\begingroup$ Have you tried with a given hyperbola and some points? $\endgroup$
    – Nosrati
    Jul 3, 2019 at 9:53
  • $\begingroup$ Where does $c^2=a^2m^2-b^2$ come from? What do $a$ and $b$ stand for? $\endgroup$ Jul 7, 2019 at 3:24
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    $\begingroup$ a=lamda ,b=lambda^3+lamda^2+lamda $\endgroup$ Jul 7, 2019 at 3:27
  • $\begingroup$ Least value of 16m^2 given in answer is 9 $\endgroup$ Jul 7, 2019 at 3:31
  • $\begingroup$ 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? $\endgroup$ Jul 7, 2019 at 3:38

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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 $\dfrac{3}{4}$ $$ \implies m^2 \ge \dfrac{9}{16} \implies 16 m^2 \ge 9 $$

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  • $\begingroup$ Sir last step how you take least is value is 3/4 for lambda quadratic equation $\endgroup$ Jul 7, 2019 at 4:13
  • $\begingroup$ 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 $\endgroup$
    – xrfxlp
    Jul 7, 2019 at 4:14

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