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Standard Equation of Parabola

The area of t...

Question

The area of the triangle formed by three points on a parabola is _____ the area of the triangle formed by the tangents at these three points.

Equal

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twice

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thrice

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four times

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Solution

The correct option is B
twice

Let's take a standard parabola $y^{2} = 4 a x$

Three points on it $P (a t_{1}^{2}, 2 a t_{1}), Q (a t_{2}^{2}, 2 a t_{2}) a n d R (a t_{3}^{2}, 2 a t_{3})$

Drawing the tangents on these three points P,Q and R

Let intersection point of tangent at P $&$ Q is A coordinates

of $A (a t_{1} t_{2}, a (t_{1} + t_{2}))$

Similarly, intersection point of tangent at Q $&$ R is B

$B (a t_{2} t_{2}, a (t_{2} + t_{3}))$

intersection point of tangent at P $&$ R is C

$(a t_{3} t_{1}, a (t_{3} + t_{1}))$

Area of triangle PQR

$\frac{1}{2} ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 a t_{1}^{2} & a t_{2}^{2} & a t_{3}^{2} 2 a t_{1} & 2 a t_{2} & 2 a t_{3} \end{matrix} ∣ ∣ ∣ ∣ ∣$

$= \frac{1}{2} [(2 a^{2} t_{2}^{2} t_{3} - 2 a^{2} t_{2} t_{3}^{2}) - (2 a^{2} t_{1}^{2} t_{3} - 2 a^{2} t_{1} t_{3}^{2}) + (2 a^{2} t_{1}^{2} t_{2} - 2 a^{2} t_{1} t_{2}^{2})]$

$= a^{2} [t_{1}^{2} (t_{2} - t_{3}) + t_{2}^{2} (t_{3} - t_{1}) + t_{3}^{2} (t_{1} - t_{3})]$

Intersection of tangents at these three points are

$[a t_{1} t_{2}, a (t_{1} + t_{2})], [a t_{2} t_{3}, a (t_{2} + t_{3})], [a t_{3} t_{1}, a (t_{3} + t_{1})]$

Area of triangle

$\frac{1}{2} ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 a t_{1} t_{2} & a t_{2} t_{3} & a t_{3} t_{1} a (t_{1} + t_{2}) & a (t_{2} + t_{3}) & a (t_{3} + t_{1}) \end{matrix} ∣ ∣ ∣ ∣$

$= \frac{1}{2} [a^{2} (t_{2} t_{3}^{2} + t_{1} t_{2} t_{3} - t_{1} t_{2} t_{3} - t_{1} t_{3}^{2}) - 1 \times a^{2} (t_{1} t_{2} t_{3} + t_{1}^{2} t_{2} - t_{1}^{2} t_{3} - t_{1} t_{2} t_{3}) + a^{2} (t_{1} t_{2}^{2} + t_{1} t_{2} t_{3} - t_{1} t_{2} t_{3} - t_{2}^{2} t_{3})]$

$= \frac{1}{2} a^{2} [t_{1}^{2} (t_{3} - t_{2}) + t_{2}^{2} (t_{1} - t_{3}) + t_{3}^{2} (t_{2} - t_{1})]$

$= \frac{- 1}{2} a^{2} [t_{1}^{2} (t_{2} - t_{3}) + t_{2}^{2} (t_{3} - t_{1}) + t_{3}^{2} (t_{1} - t_{2})]$

Area can't be negative

Taking numerical value only area $(D A B) = \frac{1}{2}$ area(DPQR)

Area of triangle formed by three points of the parabola is

twice the area of triangle formed by tangents at these three points.

Suggest Corrections

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