Question

Determine the set of values of k for which the line x + 3y = k does not intersect the curve y^2 = 2x + 3.

Answer 1

$\large\underline{\sf{Solution-}}$

Given curve is

$\rm :\longmapsto\: {y}^{2} = 2x + 3 - - - (1)$

and the equation of line is

$\rm :\longmapsto\:x + 3y = k - - - (2)$

From equation (2), we get

$\rm :\longmapsto\:x = k - 3y$

Substituting this value of x in equation (1), we get

$\rm :\longmapsto\: {y}^{2} = 2(k - 3y) + 3$

$\rm :\longmapsto\: {y}^{2} = 2k - 6y + 3$

$\rm :\longmapsto\: {y}^{2} - 2k + 6y - 3 = 0$

$\rm :\longmapsto\: {y}^{2} + 6y - 2k - 3 = 0$

So, its a quadratic equation in y.

So, for no point of intersection, Discriminant < 0

$\rm :\implies\: {b}^{2} - 4ac < 0$

Here,

$\rm :\longmapsto\:a = 1$

$\rm :\longmapsto\:b = 6$

$\rm :\longmapsto\:c = - 2k - 3$

So, on substituting the values, we get

$\rm :\longmapsto\: {6}^{2} - 4(1)( - 2k - 3) < 0$

$\rm :\longmapsto\: {6}^{2} + 4(1)(2k + 3) < 0$

$\rm :\longmapsto\: 36 + 8k + 12 < 0$

$\rm :\longmapsto\: 8k + 48< 0$

$\rm :\longmapsto\: 8k< - 48$

$\rm :\longmapsto\: k< - 6$

$\bf :\implies\:k \: \in \: ( - \: \infty , \: - 6)$

Verification :-

Let assume the value of k = - 9

So, equation of line is

$\rm :\longmapsto\:x + 3y = - 9$

On substituting x = 0, we get

$\rm :\longmapsto\:3y = - 9$

$\rm :\longmapsto\:y = - 3$

On substituting y = 0, we get

$\rm :\longmapsto\:x + 3(0) = - 9$

$\rm :\longmapsto\:x = - 9$

➢ Pair of points of the given equation are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf - 3 \\ \\ \sf - 9 & \sf 0 \end{array}} \\ \end{gathered}$

See the attachment, we concluded that line and curve donot intersect with each other.