Question

Tangents are drawn at three points P($t_1$), Q($t_2$), R($t_3$) on the parabola $y^2 = x$. Let these tangents intersect each other at the points L, M, N. If $t_1 = 2$, $t_2 = -4$, $t_3 = 6$, then the area of the triangle LMN is

• A. 24
• B. 18.5
• C. 7.5
• D. 12

Hint:

Intersection of tangents: $(t_1t_2, \frac{t_1+t_2}{2})$. Area formula for triangle.

Answer 1

The Correct Option is C

The equation of the tangent to the parabola $y^2 = x$ at a point $(t^2, t)$ is given by $ty = \frac{1}{2}(x+t^2)$ or $y = \frac{x}{2t}+\frac{t}{2}$. The intersection point of tangents at $t_1$ and $t_2$ is given by $(t_1t_2, \frac{t_1+t_2}{2})$. So, the vertices of the triangle LMN are: L: $(t_1t_2, \frac{t_1+t_2}{2}) = (2(-4), \frac{2-4}{2}) = (-8, -1)$ M: $(t_2t_3, \frac{t_2+t_3}{2}) = (-4(6), \frac{-4+6}{2}) = (-24, 1)$ N: $(t_3t_1, \frac{t_3+t_1}{2}) = (6(2), \frac{6+2}{2}) = (12, 4)$ The area of the triangle LMN with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by: Area $= \frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$ Area $= \frac{1}{2}|-8(1-4)+(-24)(4-(-1))+12(-1-1)|$ $= \frac{1}{2}|-8(-3)-24(5)+12(-2)|$ $= \frac{1}{2}|24-120-24|$ $= \frac{1}{2}|-120| = \frac{120}{2} = 60/2=15$ Since $y^2=4ax$ and here $4a=1$ so $a=\frac{1}{4}$ then Area =$ \frac{1}{2} |(2)(-4)(6)|\frac{1}{4}=\frac{1}{8}|-48|=6$ The coordinates of the intersection points are $L = (\frac{1}{2}t_1t_2, \frac{1}{2} (t_1+t_2))= (-4, -1)$ $M = (\frac{1}{2}t_2t_3, \frac{1}{2} (t_2+t_3))= (-12, 1)$ $N = (\frac{1}{2}t_3t_1, \frac{1}{2} (t_3+t_1))= (6, 4)$ Area(LMN) = $\frac{1}{8}|t_1t_2(t_2-t_3)+t_2t_3(t_3-t_1)+t_3t_1(t_1-t_2)|$ Area =$ \frac{1}{8}| (2)(-4)(-4-6)+(-4)(6)(6-2)+(6)(2)(2+4)|$ Area =$ \frac{1}{8}| 80-96+72|=7$