Question

If roots of the equation $4K^2-4K-3 = 0$ are slope of tangents drawn from a point $P$ on the parabola $y^2 = 12x$ and let tangents touch the parabola at $A$ and $B$, then which of the following is/are correct?

• A. Area of $\triangle PAB$ is $\frac{256}{3}$ square units
• B. Area of $\triangle PAB$ is $\frac{254}{3}$ square units
• C. $AB = \frac{14\sqrt{17}}{3}$ units
• D. $AB = \frac{16\sqrt{13}}{3}$ units

Answer 1

Answer: (A), (D)

A, D

Answer 2

The problem requires us to find the area of a triangle formed by a point P and the points of tangency A and B, where tangents from P are drawn to a parabola, and the length of the chord AB.

Step 1: Find the slopes of the tangents.

The given quadratic equation $4K^2-4K-3 = 0$ represents the slopes of the tangents. We solve for $K$ using the quadratic formula $K = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

$K = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(4)(-3)}}{2(4)}$

$K = \frac{4 \pm \sqrt{16 + 48}}{8}$

$K = \frac{4 \pm \sqrt{64}}{8}$

$K = \frac{4 \pm 8}{8}$

The two slopes are:

$m_1 = \frac{4 + 8}{8} = \frac{12}{8} = \frac{3}{2}$

$m_2 = \frac{4 - 8}{8} = \frac{-4}{8} = -\frac{1}{2}$

Step 2: Identify the properties of the parabola.

The equation of the parabola is $y^2 = 12x$. Comparing this with the standard form $y^2 = 4ax$, we get $4a = 12$, so $a = 3$.

Step 3: Find the point P (intersection of tangents).

The equation of a tangent to the parabola $y^2 = 4ax$ with slope $m$ is $y = mx + \frac{a}{m}$.

For $m_1 = \frac{3}{2}$ and $a=3$:

$y = \frac{3}{2}x + \frac{3}{3/2} \Rightarrow y = \frac{3}{2}x + 2$ (Equation 1)

For $m_2 = -\frac{1}{2}$ and $a=3$:

$y = -\frac{1}{2}x + \frac{3}{-1/2} \Rightarrow y = -\frac{1}{2}x - 6$ (Equation 2)

To find point P, we solve Equations 1 and 2:

$\frac{3}{2}x + 2 = -\frac{1}{2}x - 6$

$3x + 4 = -x - 12$ (multiplying by 2)

$4x = -16 \Rightarrow x = -4$

Substitute $x=-4$ into Equation 1:

$y = \frac{3}{2}(-4) + 2 = -6 + 2 = -4$

So, the point P is $(-4, -4)$.

Step 4: Find the points of tangency A and B.

The coordinates of the point of tangency for a tangent with slope $m$ to $y^2 = 4ax$ are $\left(\frac{a}{m^2}, \frac{2a}{m}\right)$.

For $m_1 = \frac{3}{2}$ and $a=3$:

Point A: $\left(\frac{3}{(3/2)^2}, \frac{2(3)}{3/2}\right) = \left(\frac{3}{9/4}, \frac{6}{3/2}\right) = \left(3 \times \frac{4}{9}, 6 \times \frac{2}{3}\right) = \left(\frac{4}{3}, 4\right)$

For $m_2 = -\frac{1}{2}$ and $a=3$:

Point B: $\left(\frac{3}{(-1/2)^2}, \frac{2(3)}{-1/2}\right) = \left(\frac{3}{1/4}, \frac{6}{-1/2}\right) = \left(3 \times 4, 6 \times (-2)\right) = \left(12, -12\right)$

So, $A = \left(\frac{4}{3}, 4\right)$ and $B = (12, -12)$.

Step 5: Calculate the area of $\triangle PAB$.

The coordinates are $P(-4, -4)$, $A(\frac{4}{3}, 4)$, $B(12, -12)$.

The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$ is $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.

Alternatively, for a parabola $y^2 = 4ax$, if tangents are drawn from an external point $P(x_1, y_1)$, the area of $\triangle PAB$ is given by $\frac{(y_1^2 - 4ax_1)^{3/2}}{2a}$.

Using this formula with $P(-4, -4)$ and $a=3$:

$y_1^2 - 4ax_1 = (-4)^2 - 4(3)(-4) = 16 + 48 = 64$

Area of $\triangle PAB = \frac{(64)^{3/2}}{2(3)} = \frac{(\sqrt{64})^3}{6} = \frac{8^3}{6} = \frac{512}{6} = \frac{256}{3}$ square units.

This matches option A.

Step 6: Calculate the length of the chord AB.

Using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ for $A(\frac{4}{3}, 4)$ and $B(12, -12)$:

$AB = \sqrt{\left(12 - \frac{4}{3}\right)^2 + (-12 - 4)^2}$

$AB = \sqrt{\left(\frac{36-4}{3}\right)^2 + (-16)^2}$

$AB = \sqrt{\left(\frac{32}{3}\right)^2 + 256}$

$AB = \sqrt{\frac{1024}{9} + 256}$

$AB = \sqrt{\frac{1024 + 256 \times 9}{9}}$

$AB = \sqrt{\frac{1024 + 2304}{9}}$

$AB = \sqrt{\frac{3328}{9}}$

To simplify $\sqrt{3328}$: $3328 = 16 \times 208 = 16 \times 16 \times 13 = 256 \times 13$.

So, $\sqrt{3328} = \sqrt{256 \times 13} = 16\sqrt{13}$.

Therefore, $AB = \frac{16\sqrt{13}}{3}$ units.

This matches option D.

The correct options are A and D.