The area of the triangle formed by the lines | vedclass

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(C) The equation of the parabola is $x^2 = 12y$. Comparing this with $x^2 = 4ay$,we get $4a = 12$,so $a = 3$.The vertex of the parabola is at $(0, 0)$

Vedclass · Accepted Answer

$18$

Vedclass · Answer

(C) The equation of the parabola is $x^2 = 12y$. Comparing this with $x^2 = 4ay$,we get $4a = 12$,so $a = 3$.The vertex of the parabola is at $(0, 0)$.The latus rectum is the line $y = a = 3$.The ends of the latus rectum are found by substituting $y = 3$ into $x^2 = 12y$,which gives $x^2 = 36$,so $x = \pm 6$.Thus,the coordinates of the ends of the latus rectum are $L_1(6, 3)$ and $L_2(-6, 3)$.The triangle is formed by the vertices $(0, 0)$,$(6, 3)$,and $(-6, 3)$.The base of the triangle is the length of the latus rectum,which is $6 - (-6) = 12$.The height of the triangle is the $y$-coordinate of the latus rectum,which is $3$.The area of the triangle is $\Delta = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 12 	imes 3 = 18 \ sq. \ unit$.

List-$I$	List-$II$
$A$. Equation of the tangent drawn at $(2, \sqrt{8})$ on the curve $y^2 = 4x$ is	$(i) -36$
$B$. Equation of the normal to the curve $y^2 = 16x$,that makes an angle of $45^{\circ}$ with its axis is	$(ii) 4$
$C$. The chord joining the points $(x_1, y_1)$ and $(x_2, y_2)$ on the curve $y^2 = 12x$ is a focal chord if $y_1 y_2 =$	$(iii) 8$
$D$. $A$ value of $k$ for which $x - 3 = 0$ is the directrix of the curve $y^2 - kx + 16 = 0$ is	$(iv) x - \sqrt{2}y + 2 = 0$
	$(v) x + y - 12 = 0$
	$(vi) x - y - 12 = 0$

The area of the triangle formed by the lines joining the vertex of the parabola $x^2 = 12y$ to the ends of its latus rectum is .................. $sq. \ unit$.

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