If y1- y2 and y3 are the ordinates of the vertices of a triangle inscribed in the parabola y2-4 a x then its area isA- 1-2 a-y1-y2y2-y3y3-y1-B- 1-8 a-y1-y2y2-y3y3-y1-C- noneD- 1-4 a-y1-y2y2-y3y3-y1-

The correct option is B Vertices of the triangle are ( y^2_1/4a,y_1 ) ( y^2_2/4a,y_2 )& ( y^2_3/4a ,y_3 ) Area of the triangle = 1/21/4a(y^2_1 y^2_2) ...

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Byju's Answer

Standard XII

Mathematics

Standard Equation of Parabola

If y1, y2 and...

Question

If $y_{1}, y_{2}$ and $y_{3}$ are the ordinates of the vertices of a triangle inscribed in the parabola $y^{2} = 4 a x$ then its area is

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Solution

The correct option is C
Vertices of the triangle are $(\frac{y_{1}^{2}}{4 a}, y_{1}) (\frac{y_{2}^{2}}{4 a}, y_{2}) & (\frac{y_{3}^{2}}{4 a}, y_{3})$
Area of the triangle = $\frac{1}{2} ∣ ∣ ∣ ∣ \begin{matrix} \frac{1}{4 a} (y_{1}^{2} - y_{2}^{2}) y_{1} - y_{2} \frac{1}{4 a} (y_{1}^{2} - y_{2}^{2}) y_{1} - y_{3} \end{matrix} ∣ ∣ ∣ ∣ = \frac{1}{2} (\frac{1}{4 a}) | (y_{1}^{2} - y_{2}^{2}) (y_{1} - y_{3}) |$
= $\frac{1}{8 a} | (y_{1} - y_{2}) (y_{2} - y_{3}) (y_{3} - y_{1}) |$

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If y1,y2 and y3 are the ordinates of the vertices of a triangle inscribed in the parabola y2=4ax then its area is

The correct option is C Vertices of the triangle are (y214a,y1)(y224a,y2)&(y234a,y3) Area of the triangle = 12∣∣ ∣∣14a(y21−y22) y1−y214a(y21−y22) y1−y3∣∣ ∣∣=12(14a)|(y21−y22)(y1−y3)| = 18a|(y1−y2)(y2−y3)(y3−y1)|

If $y_{1}, y_{2}$ and $y_{3}$ are the ordinates of the vertices of a triangle inscribed in the parabola $y^{2} = 4 a x$ then its area is