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 ______.
Correct Answer: 68
Approach Solution - 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 \(16(m_1^2 + m_2^2)\):
\[ 16(m_1^2 + m_2^2) = 16 \left( \left(-\frac{1}{2}\right)^2 + 2^2 \right). \]
\[ = 16 \left( \frac{1}{4} + 4 \right). \]
\[ = 16 \times \frac{17}{4} = 68. \]
Approach Solution -2
We are given a point \( P(0, 1) \) and a parabola \( 9x^2 + 12x + 18y - 14 = 0 \). Two tangent lines, \( L_1 \) and \( L_2 \), pass through \( P \) and touch the parabola. A triangle \( \Delta PQR \) is formed, where \( Q \) is on \( L_1 \) and \( R \) is on \( L_2 \), such that it is an isosceles triangle with base \( QR \). We need to find the value of \( 16\left(m_1^2 + m_2^2\right) \), where \( m_1 \) and \( m_2 \) are the possible slopes of the line \( QR \).
Concept Used:
1. Standard Equation of a Parabola: To analyze the parabola, we first convert its equation into the standard form \( (x-h)^2 = 4a(y-k) \) or \( (y-k)^2 = 4a(x-h) \).
2. Condition for Tangency: A line \( y = mx + c \) is tangent to a parabola if the quadratic equation formed by substituting \( y \) into the parabola's equation has a single, repeated root. This means the discriminant (\( D \)) of the quadratic equation must be zero.
3. Geometry of Isosceles Triangles: In an isosceles triangle \( \Delta PQR \) with base \( QR \), the altitude from the vertex \( P \) to the base \( QR \) is also the angle bisector of the angle \( \angle QPR \).
4. Angle Bisectors of Lines: The equations of the angle bisectors of two lines \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \) are given by \( \frac{a_1x + b_1y + c_1}{\sqrt{a_1^2 + b_1^2}} = \pm \frac{a_2x + b_2y + c_2}{\sqrt{a_2^2 + b_2^2}} \).
5. Perpendicular Lines: If two lines have slopes \( m_a \) and \( m_b \), they are perpendicular if and only if \( m_a \cdot m_b = -1 \).
Step-by-Step Solution:
Step 1: Convert the equation of the parabola to its standard form.
\[ 9x^2 + 12x + 18y - 14 = 0 \]Complete the square for the \( x \) terms:
\[ 9(x^2 + \frac{4}{3}x) = -18y + 14 \] \[ 9(x^2 + \frac{4}{3}x + \frac{4}{9}) = -18y + 14 + 9(\frac{4}{9}) \] \[ 9(x + \frac{2}{3})^2 = -18y + 18 \] \[ 9(x + \frac{2}{3})^2 = -18(y - 1) \] \[ (x + \frac{2}{3})^2 = -2(y - 1) \]This is a parabola with its vertex at \( (-2/3, 1) \) and opening downwards.
Step 2: Find the slopes of the tangents \( L_1 \) and \( L_2 \) from the point \( P(0, 1) \).
The equation of any line passing through \( P(0, 1) \) with slope \( s \) is \( y - 1 = s(x - 0) \), which simplifies to \( y = sx + 1 \).
Substitute this into the parabola's equation:
\[ (x + \frac{2}{3})^2 = -2((sx + 1) - 1) \] \[ x^2 + \frac{4}{3}x + \frac{4}{9} = -2sx \] \[ x^2 + (\frac{4}{3} + 2s)x + \frac{4}{9} = 0 \]For this line to be a tangent, the discriminant of this quadratic equation must be zero.
\[ D = (\frac{4}{3} + 2s)^2 - 4(1)(\frac{4}{9}) = 0 \] \[ (2(s + \frac{2}{3}))^2 = \frac{16}{9} \] \[ 4(s + \frac{2}{3})^2 = \frac{16}{9} \implies (s + \frac{2}{3})^2 = \frac{4}{9} \]Taking the square root of both sides gives:
\[ s + \frac{2}{3} = \pm \frac{2}{3} \]This gives two possible slopes for the tangents:
\[ s_1 = \frac{2}{3} - \frac{2}{3} = 0 \] \[ s_2 = -\frac{2}{3} - \frac{2}{3} = -\frac{4}{3} \]Step 3: Determine the equations of the tangent lines \( L_1 \) and \( L_2 \).
For \( s_1 = 0 \): \( L_1: y - 1 = 0 \).
For \( s_2 = -4/3 \): \( L_2: y - 1 = -\frac{4}{3}x \implies 4x + 3y - 3 = 0 \).
Step 4: Find the slopes of the angle bisectors of the angle between \( L_1 \) and \( L_2 \).
Since \( \Delta PQR \) is isosceles with base \( QR \), the altitude from \( P \) to \( QR \) must be an angle bisector of \( \angle QPR \). The line \( QR \) is therefore perpendicular to this angle bisector.
The equations of the angle bisectors are:
\[ \frac{y - 1}{\sqrt{0^2 + 1^2}} = \pm \frac{4x + 3y - 3}{\sqrt{4^2 + 3^2}} \] \[ y - 1 = \pm \frac{4x + 3y - 3}{5} \]Bisector 1: \( 5(y - 1) = 4x + 3y - 3 \implies 5y - 5 = 4x + 3y - 3 \implies 2y = 4x + 2 \implies y = 2x + 1 \). The slope is \( m_{b1} = 2 \).
Bisector 2: \( 5(y - 1) = -(4x + 3y - 3) \implies 5y - 5 = -4x - 3y + 3 \implies 8y = -4x + 8 \implies y = -\frac{1}{2}x + 1 \). The slope is \( m_{b2} = -1/2 \).
Step 5: Find the possible slopes \( m_1 \) and \( m_2 \) of the line \( QR \).
The line \( QR \) must be perpendicular to one of the angle bisectors. Thus, the slopes \( m_1 \) and \( m_2 \) are the negative reciprocals of the slopes of the bisectors.
\[ m_1 = -\frac{1}{m_{b1}} = -\frac{1}{2} \] \[ m_2 = -\frac{1}{m_{b2}} = -\frac{1}{-1/2} = 2 \]Final Computation & Result:
Step 6: Calculate the value of \( 16(m_1^2 + m_2^2) \).
\[ m_1^2 = (-\frac{1}{2})^2 = \frac{1}{4} \] \[ m_2^2 = (2)^2 = 4 \]The sum of the squares is:
\[ m_1^2 + m_2^2 = \frac{1}{4} + 4 = \frac{1 + 16}{4} = \frac{17}{4} \]Finally, multiply by 16:
\[ 16(m_1^2 + m_2^2) = 16 \times \frac{17}{4} = 4 \times 17 = 68 \]The value of \( 16\left(m_1^2 + m_2^2\right) \) is 68.
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