Question

Let $L_1, L_2$ be the lines passing through the point $P(0, 1)$ and touching the parabola $9x^2 + 12x + 18y - 14 = 0.$ Let $Q$ and $R$ be the points on the lines $L_1$ and $L_2$ such that the $\Delta PQR$ is an isosceles triangle with base $QR$. If the slopes of the lines $QR$ are $m_1$ and $m_2$, then $16\left(m_1^2 + m_2^2\right)$ is equal to ______.

Answer 1

The given parabola is:

$9x^2 + 12x + 4 = -18(y - 1).$

Rewriting:

$(3x + 2)^2 = -18(y - 1).$

Equation of a line passing through $P(0, 1)$ is:

$y = mx + 1.$

Substituting $y = mx + 1$ into the parabola:

$(3x + 2)^2 = -18(mx).$

Expanding:

$9x^2 + (12 + 18m)x + 4 = 0.$

For the line to be tangent to the parabola:

$\Delta = 0.$

Using the discriminant condition:

The equation simplifies as:

$(12 + 18m)^2 - 4(9)(4) = 0.$

Simplify:

$144 + 432m + 324m^2 - 144 = 0.$

$36m^2 + 108m + 36 = 0.$

Simplify:

$m^2 + 3m + 1 = 0.$

Solving the quadratic equation:

$m = \frac{-3 \pm \sqrt{3^2 - 4(1)(1)}}{2(1)} = \frac{-3 \pm \sqrt{5}}{2}.$

This gives the slopes $m_1 = \frac{-3 + \sqrt{5}}{2}$ and $m_2 = \frac{-3 - \sqrt{5}}{2}$.

Step 2: Finding slopes of $\overline{QR}$:

For $\triangle PQR$, the triangle is isosceles with base $\overline{QR}$. Using the tangent property, the slope of $\overline{QR}$ can be derived using:

$\text{Slope of QR} = \tan \left( \frac{\pi}{2} + \frac{\theta}{2} \right).$

Using symmetry, calculate $\theta$:

$\theta = \tan^{-1} \left( \frac{4}{3} \right).$

Then:

$m_1 = -\cot\left(\frac{\theta}{2}\right), \quad m_2 = \tan\left(\frac{\theta}{2}\right).$

Finally, calculate:

$m_1 = -\frac{1}{2}, \quad m_2 = 2.$

Step 3: Calculating $16(m_1^2 + m_2^2)$:

$16(m_1^2 + m_2^2) = 16 \left( \left(-\frac{1}{2}\right)^2 + 2^2 \right).$

$= 16 \left( \frac{1}{4} + 4 \right).$

$= 16 \times \frac{17}{4} = 68.$