Triangle ABC is inscribed in the parabola described by the equation y 2-6 x -4 y +10=0 so that A is the vertex of the parabola and B and C are the end points of the latus rectum of the parabola. The area of triangle A B C is

Question

Triangle ABC is inscribed in the parabola described by the equation y 2-6 x -4 y +10=0 so that A is the vertex of the parabola and B and C are the end points of the latus rectum of the parabola. The area of triangle A B C is (A) 18 (B) 9 (C) 4.5 (D) 2.25

Tardigrade Admin · Accepted Answer

(y-2)2-4=6(x-(5/3))=6 x-10 (y-2)2=6 x-6=6(x-1) Let y-2=Y and x-1=X ∴ Y 2=6 X Area of triangle ABC, =(3/2) ⋅ 6 ⋅ (1/2)=(9/2)

Q. Triangle $A BC$ is inscribed in the parabola described by the equation $y^{2} - 6 x - 4 y + 10 = 0$ so that $A$ is the vertex of the parabola and $B$ and $C$ are the end points of the latus rectum of the parabola. The area of triangle $A BC$ is

18

9

4.5

2.25

$(y - 2)^{2} - 4 = 6 (x - \frac{5}{3}) = 6 x - 10$
$(y - 2)^{2} = 6 x - 6 = 6 (x - 1)$
Let $y - 2 = Y$ and $x - 1 = X$
$∴ Y^{2} = 6 X$
Area of $△ A BC$ ,
$= \frac{3}{2} \cdot 6 \cdot \frac{1}{2} = \frac{9}{2}$

Q. Triangle ABC is inscribed in the parabola described by the equation y2−6x−4y+10=0 so that A is the vertex of the parabola and B and C are the end points of the latus rectum of the parabola. The area of triangle ABC is

18

9

4.5

2.25

(y−2)2−4=6(x−35​)=6x−10 (y−2)2=6x−6=6(x−1) Let y−2=Y and x−1=X ∴Y2=6X Area of △ABC, =23​⋅6⋅21​=29​

Q. Triangle $A BC$ is inscribed in the parabola described by the equation $y^{2} - 6 x - 4 y + 10 = 0$ so that $A$ is the vertex of the parabola and $B$ and $C$ are the end points of the latus rectum of the parabola. The area of triangle $A BC$ is

$(y - 2)^{2} - 4 = 6 (x - \frac{5}{3}) = 6 x - 10$
$(y - 2)^{2} = 6 x - 6 = 6 (x - 1)$
Let $y - 2 = Y$ and $x - 1 = X$
$∴ Y^{2} = 6 X$
Area of $△ A BC$ ,
$= \frac{3}{2} \cdot 6 \cdot \frac{1}{2} = \frac{9}{2}$