Question

Consider the parabola ${y^2} = 8x$. Let ${\vartriangle _1}$ be the area of the triangle formed by the end points of its latus rectum and the point $P\left( {\dfrac{1}{2},2} \right)$ on the parabola, and ${\vartriangle _2}$ be the area of the triangle formed by drawing tangents at $P$ and the end points of the latus rectum. Then $\dfrac{{{\vartriangle _1}}}{{{\vartriangle _2}}}$is :
(A) $2$
(B) $3$
(C) $4$
(D) $1$

Answer 1

Hint: Firstly, find out the endpoints of latus rectum of the given parabola ${y^2} = 8x$ ,i.e., $\left( {a,2a} \right)$and $\left( {a, - 2a} \right)$. Now, find the equations of tangent $\left[ {y - {y_1} = \left( {\dfrac{{dy}}{{dx}}} \right)\left( {x - {x_1}} \right)} \right]$ at point $P$ and the end points of the latus rectum and solve them to find the point of intersection.
Find the area of triangle formed by joining the three given points $\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right)$ and $\left( {{x_3},{y_3}} \right)$ by using the formula :$\dfrac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]$.

Complete step-by-step answer:
Given, Equation of parabola ${y^2} = 8x$
Compare this equation of parabola with the standard equation of parabola i.e., ${y^2} = 4ax$, we get-
$4a = 8 \Rightarrow a = 2$ …. (1)
The end points of latus rectum are given by $M$$\left( {a,2a} \right)$ and $N$$\left( {a, - 2a} \right)$. So end points of latus rectum are $M$$\left( {2,4} \right)$ and $N$$\left( {2, - 4} \right)$.
The area of triangle formed by joining the three given points is calculated by $\dfrac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]$.
So, the area of triangle formed by joining the end points of latus rectum $M$$\left( {2,4} \right)$,$N$$\left( {2, - 4} \right)$and $P\left( {\dfrac{1}{2},2} \right)$
${\vartriangle _1}$$ = \dfrac{1}{2}\left[ {2\left( { - 4 - 2} \right) + 2\left( {2 - 4} \right) + \dfrac{1}{2}\left( {4 + 4} \right)} \right]$
${\vartriangle _1}$$ = \dfrac{1}{2}\left[ {2\left( { - 6} \right) + 2\left( { - 2} \right) + \dfrac{1}{2}\left( 8 \right)} \right]$
${\vartriangle _1}$$ = \dfrac{1}{2}\left[ { - 12 - 4 + 4} \right]$
${\vartriangle _1}$$ = \dfrac{1}{2} \times - 12$
${\vartriangle _1}$$ = - 6$
Since, the area of a triangle can never be negative.
Therefore, ${\vartriangle _1}$$ = 6$ sq. units
Now we have to find ${\vartriangle _2}$ which is the area of the triangle formed by drawing tangents at $P$ and the end points of the latus rectum $M$ and $N$.
Given, Equation of parabola ${y^2} = 8x$
Differentiate it with respect to $x$ to find out the value of slope, i.e., $m = $$\dfrac{{dy}}{{dx}}$
$
   \Rightarrow 2y\dfrac{{dy}}{{dx}} = 8 \\
   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{8}{{2y}} \\
   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{4}{y} \\
 $
Thus, $m = \dfrac{{dy}}{{dx}} = \dfrac{4}{y}$
Slope at point $M\left( {2,4} \right)$$ \Rightarrow $$m = $${\left( {\dfrac{{dy}}{{dx}}} \right)_M} = \dfrac{4}{4} = 1$
The equation of tangent at $M\left( {2,4} \right)$ is given by
$y - {y_1} = {\left( {\dfrac{{dy}}{{dx}}} \right)_M}\left( {x - {x_1}} \right)$
$ \Rightarrow y - 4 = 1\left( {x - 2} \right)$
$ \Rightarrow y - 4 = x - 2$
$ \Rightarrow y = x + 2$ …. (1)
Slope at point $N\left( {2, - 4} \right)$$ \Rightarrow $$m = $${\left( {\dfrac{{dy}}{{dx}}} \right)_N} = \dfrac{4}{{ - 4}} = - 1$
The equation of tangent at $N\left( {2, - 4} \right)$ is given by
$y - {y_1} = {\left( {\dfrac{{dy}}{{dx}}} \right)_N}\left( {x - {x_1}} \right)$
$ \Rightarrow y + 4 = - 1\left( {x - 2} \right)$
$ \Rightarrow y + 4 = - x + 2$
$ \Rightarrow y = - x - 2$…. (2)
Slope at point $P\left( {\dfrac{1}{2},2} \right)$$ \Rightarrow $$m = $${\left( {\dfrac{{dy}}{{dx}}} \right)_P} = \dfrac{4}{2} = 2$
The equation of tangent at$P\left( {\dfrac{1}{2},2} \right)$ is given by
$y - {y_1} = {\left( {\dfrac{{dy}}{{dx}}} \right)_P}\left( {x - {x_1}} \right)$
$ \Rightarrow y - 2 = 2\left( {x - \dfrac{1}{2}} \right)$
$ \Rightarrow y - 2 = 2x - 1$
$ \Rightarrow y = 2x + 1$ …. (3)
On solving (1) and (2), we get the point of intersection of (1) and (2), i.e., $\left( { - 2,0} \right) & $.
On solving (2) and (3), we get the point of intersection of (2) and (3), i.e., $\left( { - 1, - 1} \right)$ .
On solving (1) and (3), we get the point of intersection of (1) and (3), i.e., $\left( {1,3} \right)$.
Now ${\vartriangle _2}$is the area of triangle formed by joining the points $\left( { - 2,0} \right),\left( { - 1, - 1} \right)$and $\left( {1,3} \right)$
${\vartriangle _2}$=$\dfrac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]$
$ \Rightarrow $${\vartriangle _2} = \dfrac{1}{2}\left[ { - 2\left( { - 1 - 3} \right) - 1\left( {3 - 0} \right) + 1\left( {0 + 1} \right)} \right]$
$ \Rightarrow $${\vartriangle _2} = \dfrac{1}{2}\left[ { - 2\left( { - 4} \right) - 1\left( 3 \right) + 1\left( 1 \right)} \right]$
$ \Rightarrow $${\vartriangle _2} = \dfrac{1}{2}\left[ {8 - 3 + 1} \right]$
$ \Rightarrow $\[{\vartriangle _2} = \dfrac{1}{2} \times 6\]
$ \Rightarrow $\[{\vartriangle _2} = 3\] sq. units
Now, $\dfrac{{{\vartriangle _1}}}{{{\vartriangle _2}}} = \dfrac{6}{3} = 2$

Hence, option (A) is the correct answer.

Note: An alternative method to solve this question is a theorem which states that the area of a triangle formed by three points on a parabola is twice the area of triangle formed by tangents at those points.
$\therefore {\vartriangle _1} = 2{\vartriangle _2}$$ \Rightarrow \dfrac{{{\vartriangle _1}}}{{{\vartriangle _2}}} = 2$