Prove that the area of the triangle formed by three points on a parabola is twice the area of the triangle formed by the tangents at these points.

Open in App

Solution

Equation of the parabola is $y^{2} = 4 a x$
Let $A (a t_{1}^{2}, 2 a t_{1}), B (a t_{2}^{2}, 2 a t_{2})$ and $C (a t_{3}^{2}, 2 a t_{3})$ be the three given points on the given parabola.
Equation of the tangent to the parabola at $A (a t_{1}^{2}, 2 a t_{1})$ is
$y t_{1} = x + a t_{1}^{2}$ ........ $(1)$
Equation of tangent to the parabola at $B (a t_{2}^{2}, 2 a t_{2})$ is
$y t_{2} = x + a t_{2}^{2}$ ........ $(2)$
and $C (a t_{3}^{2}, 2 a t_{3})$ is
$y t_{3} = x + a t_{3}^{2}$ ........ $(3)$
Suppose the tangents at $A, B$ and $C$ intersect at $P, Q$ and $R$ respectively.
Solving $(1)$ and $(2)$ we get
$x = a t_{1} t_{2}$ and $y = a (t_{1} + t_{2})$
$∴$ Co-ordinate of $P = (a t_{1} t_{2}, a (t_{1} + t_{2}))$
Solving $(2)$ and $(3)$ we get
$x = a t_{2} t_{3}$ and $y = a (t_{2} + t_{3})$
$∴$ Co-ordinate of $Q = (a t_{2} t_{3}, a (t_{2} + t_{3}))$
Solving $(3)$ and $(1)$ we get
$x = a t_{3} t_{1}$ and $y = a (t_{3} + t_{1})$
$∴$ Co-ordinate of $Q = (a t_{3} t_{1}, a (t_{3} + t_{1}))$
Area $△ A B C = \frac{1}{2} ∣ ∣ a t_{1}^{2} (2 a t_{1} - 2 a t_{2}) + a t_{2}^{2} (2 a t_{3} - 2 a t_{1}) + a t_{3}^{2} (2 a t_{1} - 2 a t_{2}) ∣ ∣$
This expression on simplification gives
Area of $△ A B C = \frac{1}{2} a^{2} | (t_{1} - t_{2}) (t_{2} - t_{3}) (t_{3} - t_{1}) |$
Area of $△ P Q R = \frac{1}{2} | a t_{1} t_{2} [a (t_{2} + t_{3}) - a (t_{1} + t_{3})] + a t_{2} t_{3} [a (t_{1} + t_{3}) - a (t_{1} + t_{2})] + a t_{3} t_{1} [a (t_{1} + t_{2}) - a (t_{2} + t_{3})] |$
This expression on simplification gives
Area of $△ P Q R = \frac{1}{2} \times \frac{a^{2}}{2} | (t_{1} - t_{2}) (t_{2} - t_{3}) (t_{3} - t_{1}) |$
$\frac{A r e a (△ A B C)}{A r e a (△ P Q R)} = \frac{\frac{a^{2}}{2} | (t_{1} - t_{2}) (t_{2} - t_{3}) (t_{3} - t_{1}) |}{\frac{a^{2}}{4} | (t_{1} - t_{2}) (t_{2} - t_{3}) (t_{3} - t_{1}) |} = 2$
$∴$ Area of $(△ A B C) = 2 \times$ Area of $(△ P Q R)$

Suggest Corrections

Similar questions

y^{2} = 8 x

Δ_{1}

P (\frac{1}{2}, 2)

Δ_{2}

P

\frac{Δ_{1}}{Δ_{2}}

t_{1}, t_{2}, t_{3}

y^{2} = 4 a x

y^{2} = 8 x

Δ_{1}

P (\frac{1}{2}, 2)

Δ_{2}

\frac{Δ_{1}}{Δ_{2}}

(a t_{1}^{2}, 2 a t_{1}), (a t_{2}^{2}, 2 a t_{2})

(a t_{3}^{2}, 2 a t_{3})

\frac{a^{2}}{2} (t_{2} - t_{3}) (t_{3} - t_{1}) (t_{1} - t_{2}) (t_{1} + t_{2} + t_{3})^{2} .

Join BYJU'S Learning Program