The ratio of the area of a triangle inscribed in a parabola to the area of the triangle formed by the tangents drawn at the vertices of the triangle is

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Solution

Let the parabola be $y^{2} = 4 a x$
Assuming the points of the triangle be $A, B, C$ as
$(a t_{1}^{2}, 2 a t_{1}), (a t_{2}^{2}, 2 a t_{2}), (a t_{3}^{2}, 2 a t_{3})$ respectively.

Let $P, Q, R$ be the points of intersection of tangents at $B, C; C, A; A, B$
$∴ P = (a t_{2} t_{3}, a (t_{2} + t_{3}))$
Similarly,
$Q = (a t_{1} t_{3}, a (t_{1} + t_{3})) R = (a t_{2} t_{1}, a (t_{2} + t_{1}))$

Area of
$△ A B C = \frac{1}{2} ∣ ∣ ∣ ∣ ∣ \begin{matrix} a t_{1}^{2} & 2 a t_{1} & 1 a t_{2}^{2} & 2 a t_{2} & 1 a t_{3}^{2} & 2 a t_{3} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ applying R_{1} \to R_{1} - R_{3} R_{2} \to R_{2} - R_{3} = a^{2} ∣ ∣ ∣ ∣ ∣ \begin{matrix} t_{1}^{2} - t_{3}^{2} & t_{1} - t_{3} & 0 t_{2}^{2} - t_{3}^{2} & t_{2} - t_{3} & 0 t_{3}^{2} & t_{3} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = a^{2} | (t_{1} - t_{2}) (t_{2} - t_{3}) (t_{1} - t_{3}) |$

Area of
$△ P Q R = \frac{1}{2} ∣ ∣ ∣ ∣ ∣ \begin{matrix} a t_{3} t_{2} & a (t_{2} + t_{3}) & 1 a t_{3} t_{1} & a (t_{3} + t_{1}) & 1 a t_{1} t_{2} & a (t_{1} + t_{2}) & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ applying R_{1} \to R_{1} - R_{2} R_{2} \to R_{2} - R_{3} = \frac{a^{2}}{2} ∣ ∣ ∣ ∣ ∣ \begin{matrix} t_{3} (t_{2} - t_{1}) & (t_{2} - t_{1}) & 0 t_{1} (t_{3} - t_{2}) & (t_{3} - t_{2}) & 0 t_{1} t_{2} & (t_{1} + t_{2}) & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = \frac{a^{2}}{2} | (t_{1} - t_{2}) (t_{2} - t_{3}) (t_{1} - t_{3}) | \Rightarrow 2 △ P Q R = △ A B C$

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