The area of the triangle formed by the lines | vedclass

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(B) The equation of the parabola is $x^2 = 48y$. Comparing this with $x^2 = 4ay$, we get $4a = 48$, so $a = 12$.The vertex of the parabola is $V(0, 0)

Vedclass · Accepted Answer

$288 	ext{ sq. units}$

Vedclass · Answer

(B) The equation of the parabola is $x^2 = 48y$. Comparing this with $x^2 = 4ay$, we get $4a = 48$, so $a = 12$.The vertex of the parabola is $V(0, 0)$.The ends of the latus rectum are at $(2a, a)$ and $(-2a, a)$, which are $(24, 12)$ and $(-24, 12)$.The area of the triangle formed by vertices $(0, 0)$, $(24, 12)$, and $(-24, 12)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.Area $= \frac{1}{2} |0(12 - 12) + 24(12 - 0) + (-24)(0 - 12)|$.Area $= \frac{1}{2} |0 + 288 + 288| = \frac{1}{2} |576| = 288 	ext{ sq. units}$.

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 $x^2 = 48y$ to the ends of its latus rectum is .....

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