Question

If the tangents of the parabola $ y^2 = 8x $ passing through the point $ P(1, 3) $ touch the parabola at points $ A $ and $ B $, then the area (in sq. units) of $ \triangle ABC $ is

• A. $ 1 $
• B. $ \frac{3}{4} $
• C. $ \frac{1}{2} $
• D. $ \frac{1}{4} $

Hint:

To find the area of a triangle formed by tangents to a parabola, use the chord of contact to determine the points of contact and then apply the formula for the area of a triangle.

Answer 1

The Correct Option is D

Step 1: Equation of the tangent to the parabola. 
The given parabola is: 
$$ y^2 = 8x. $$ The general equation of a tangent to the parabola \(y^2 = 4ax\) at a point \((x_1, y_1)\) is: 
$$ yy_1 = 2a(x + x_1). $$ For \(y^2 = 8x\), we have \(4a = 8 \Rightarrow a = 2\). Thus, the tangent equation becomes: 
$$ yy_1 = 4(x + x_1). $$ Step 2: Tangents passing through \(P(1, 3)\). 
Let the tangents pass through \(P(1, 3)\). Substituting \((x, y) = (1, 3)\) into the tangent equation: 
$$ 3y_1 = 4(1 + x_1). $$ Rearrange: 
$$ 3y_1 = 4 + 4x_1 \quad \Rightarrow \quad 4x_1 - 3y_1 + 4 = 0. $$ This is the chord of contact of the tangents from \(P(1, 3)\) to the parabola. The points of contact \(A\) and \(B\) lie on this line. 
Step 3: Points of intersection of the chord of contact with the parabola. 
Substitute \(y_1 = \frac{4x_1 + 4}{3}\) into the parabola equation \(y^2 = 8x\): 
$$ \left(\frac{4x_1 + 4}{3}\right)^2 = 8x_1. $$ Simplify: 
$$ \frac{(4x_1 + 4)^2}{9} = 8x_1 \quad \Rightarrow \quad (4x_1 + 4)^2 = 72x_1. $$ Expand: 
$$ 16x_1^2 + 32x_1 + 16 = 72x_1 \quad \Rightarrow \quad 16x_1^2 - 40x_1 + 16 = 0. $$ Divide by 4: 
$$ 4x_1^2 - 10x_1 + 4 = 0. $$ Solve using the quadratic formula: 
$$ x_1 = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(4)(4)}}{2(4)} = \frac{10 \pm \sqrt{100 - 64}}{8} = \frac{10 \pm \sqrt{36}}{8} = \frac{10 \pm 6}{8}. $$ Thus: 
$$ x_1 = \frac{16}{8} = 2 \quad \text{or} \quad x_1 = \frac{4}{8} = \frac{1}{2}. $$ Corresponding \(y_1\)-coordinates are: 
$$ y_1 = \frac{4x_1 + 4}{3}. $$ For \(x_1 = 2\): 
$$ y_1 = \frac{4(2) + 4}{3} = \frac{8 + 4}{3} = 4. $$ For \(x_1 = \frac{1}{2}\): 
$$ y_1 = \frac{4\left(\frac{1}{2}\right) + 4}{3} = \frac{2 + 4}{3} = 2. $$ Thus, the points of contact are: 
$$ A(2, 4) \quad \text{and} \quad B\left(\frac{1}{2}, 2\right). $$ Step 4: Area of \( \triangle ABC \). 
The vertices of \( \triangle ABC \) are \( A(2, 4) \), \( B\left(\frac{1}{2}, 2\right) \), and \( C(1, 3) \). The area of a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) is given by: 
$$ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|. $$ Substitute \( A(2, 4) \), \( B\left(\frac{1}{2}, 2\right) \), and \( C(1, 3) \): 
$$ \text{Area} = \frac{1}{2} \left| 2(2 - 3) + \frac{1}{2}(3 - 4) + 1(4 - 2) \right|. $$ Simplify: 
$$ \text{Area} = \frac{1}{2} \left| 2(-1) + \frac{1}{2}(-1) + 1(2) \right| = \frac{1}{2} \left| -2 - \frac{1}{2} + 2 \right| = \frac{1}{2} \left| -\frac{1}{2} \right| = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}. $$ Step 5: Final Answer. 
$$ \boxed{\frac{1}{4}} $$