Question #209e6

Sep 13, 2016

See the Explanation.

Explanation:

Suppose, to the contrary , that

$\exists \text{ a line, say "l," with Y-intercept "10" and touching the Curve}$

$\text{(Parabola) C : } y = 3 {x}^{2} + 7 x - 2$.

Since, $l$ touches $C$, $l \cap C$ must be a Singleton $\subset {\mathbb{R}}^{2}$.

If the slope of $l$ is $m$, then, the eqn. of $l$ is $y = m x + 10$.

[A Clarification : In case, $m$ does not exist, then, $l$ has to be

vertical, i.e., l || Y-Axis; so, l does not intersect Y-Axis, &, as such,

$l \text{ can not have "Y"-intercept"=10". Evidently, } m$ does exist. ]

To find $l \cap C$, we solve their eqns.

$y = m x + 10 , y = 3 {x}^{2} + 7 x - 2 \Rightarrow m x + 10 = 3 {x}^{2} + 7 x - 2$.

$\therefore 3 {x}^{2} + \left(7 - m\right) x - 12 = 0. \ldots \ldots \ldots \ldots \ldots . . \left(\star\right)$

In order that $l \cap C$ be Singleton, the qudr. eqn.$\left(\star\right)$ must

have two identical roots, for which $\Delta = 0$.

$\therefore {\left(7 - m\right)}^{2} - 4 \left(3\right) \left(- 12\right) = 0$

${\left(7 - m\right)}^{2} = - 144$, Impossible in $\mathbb{R}$.

This contradiction shows that our supposition is wrong.

Thus, no line having $Y$-intercept $10$ can be tangential to

the given Parabola.

Enjoy Maths.!