Question

If \(\mathrm{y}=-2\) is a tangent \(\Rightarrow \mathrm{px}^{2}+\mathrm{qx}+\mathrm{r}+2=0 \quad\) will have only one real root \(\Rightarrow \mathrm{q}^{2}-4 \mathrm{p}(\mathrm{r}+2)=0\) \(=\mathrm{q}^{2}-4 \mathrm{pr}-8 \mathrm{p}=0\) \(\Rightarrow \mathrm{p}<0\) as \(\left(\mathrm{q}^{2}-4 \mathrm{pr}<0\right)\) \(\Rightarrow \mathrm{r}<0\)

Short answer

Answer: The line y = -2 will be a tangent to the quadratic equation when p < 0 and r > - (q^2 / 4p) - 2.

Step-by-step solution

Set the discriminant equal to zero

To find the quadratic equation with one real root, we need to set the discriminant equal to zero. So, \(\mathrm{q}^{2} - 4\mathrm{p}(\mathrm{r} + 2) = 0\).

Simplify the equation

Next, we simplify the equation by expanding. This gives us: \(\Rightarrow \mathrm{q}^{2} - 4\mathrm{pr} - 8\mathrm{p} = 0\)

Determine the inequality for p

Recall that discriminant is positive if the quadratic equation has two distinct real roots, negative if the equation has two imaginary roots, and zero if the quadratic equation has exactly one real root (a tangent). Since we want \(\mathrm{q}^{2} - 4\mathrm{pr} < 0\), we start by observing that: $$ \mathrm{p} < 0 $$

Determine the inequality for r

Now, to find the inequality for r, we can plug p back into the equation, \(\mathrm{q}^2 - 4\mathrm{pr} - 8\mathrm{p} = 0\). Since we know that \(\mathrm{p}<0\), we can divide the equation by \(-4\mathrm{p}\). Doing this gives: $$ -\frac{\mathrm{q}^2}{4\mathrm{p}} + \mathrm{r} + 2 > 0 $$ Rearrange the inequality to isolate r: $$ \Rightarrow \mathrm{r} > -\frac{\mathrm{q}^2}{4\mathrm{p}} - 2 $$