Problem to make you happy

From: $3x+5y+n=0\quad \Rightarrow x = -\dfrac {5y+n}{3}$ .

$\Rightarrow y^2 = 24x = 24\left(-\dfrac {5y+n}{3}\right) = -40y-8n\quad \Rightarrow y^2+40y+8n = 0$

For the line to be tangential to the parabola, the above equation has only one root and this occurs when:

$\quad 40^2 - 4(8n) = 0\quad \Rightarrow n = \dfrac {40\times 40}{4\times 8} = \boxed{50}$

Equation of the tangent to the parabola $\displaystyle { y }^{ 2 }=4ax$ is in the form $\displaystyle y=mx+\cfrac { a }{ m }$

Therefore, equation of tangent to the parabola $\displaystyle { y }^{ 2 }=4.6.x$ will be of the form $\displaystyle y=mx+\cfrac { 6 }{ m }$ . Since the value of $\displaystyle m=-\cfrac { 3 }{ 5 }$ in the equation $\displaystyle 3x+5y+n=0$ , value of $\displaystyle n$ can be found as follows;

$\displaystyle 6\times \left( -\cfrac { 5 }{ 3 } \right) =\left( -\cfrac { n }{ 5 } \right) \Rightarrow n=50$

From : 3x+5y+n=0 we can write in standard form of line equation i.e. y=(-3/5)x+(-1/5)n; so the slope of the line is -3/5, which must be equal to the slope from parabola as the line is tangent. So, dy/dx= 12/y. Now, equating the slopes we get y= -20. put this in the parabola and we get x= 50/3. Now, put these value in the equation of line as line and parabola share common point at this, hence n=50.

Put x=y^2/24 in the tangent equation. Equate the discriminant, D=0. 1600-32n=0. n=50