Parabola Ques 12
- Points $A, B$ and $C$ lie on the parabola $y^2=4 a x$. The tangents to the parabola at $A, B$ and $C$, taken in pairs, intersect at points $P, Q$ and $R$. Determine the ratio of the areas of the triangles $A B C$ and $P Q R$.
$(1996,3 M)$
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Answer:
Correct Answer: 12.(2)
Solution: Let the three points on the parabola be $A\left(a t_1^2, 2 a t_1\right), B\left(a t_2^2, 2 a t_2\right)$ and $C\left(a t_3^2, 2 a t_3\right)$
Equation of the tangent to the parabola at $\left(a t^2, 2 a t\right)$ is
$ t y=x+a t^2 $
Therefore, equations of tangents at $A$ and $B$ are
$ \begin{aligned} & t_1 y=x+a t_1^2 \quad ……..(i) \\ & t_2 y=x+a t_2^2 \quad ……..(i) \end{aligned} $
From Eqs. (i) and (ii)
$ \begin{aligned} & & t_1 y & =t_2 y-a t_2^2+a t_1^2 \\ \Rightarrow & & t_1 y-t_2 y & =a t_1^2-a t_2^2 \\ \Rightarrow & & y & =a\left(t_1+t_2\right) \\ \Rightarrow & & t_1 a\left(t_1+t_2\right) & =x+a t_1^2 \quad \text{[from Eq. (i)]} \\ & & x & =a t_1 t_2 \end{aligned} $
Therefore, coordinates of $P$ are $\left(a t_1 t_2, a\left(t_1+t_2\right)\right)$.
Similarly, the coordinates of $Q$ and $R$ are respectively,
$ \left[a t_2 t_3, a\left(t_2+t_3\right)\right] \text { and }\left[a t_1 t_3, a\left(t_1+t_3\right)\right] \text {. } $
Let $\quad \Delta_1=$ Area of the $\triangle A B C$
$=\frac{1}{2}\left|\begin{array}{lll} a t_1^2 & 2 a t_1 & 1 \\ a t_2^2 & 2 a t_2 & 1 \\ a t_3^2 & 2 a t_3 & 1 \end{array}\right|$
Applying $R_3 \rightarrow R_3-R_2$ and $R_2 \rightarrow R_2-R_1$, we get
$ \Delta_1 =\frac{1}{2}\begin{vmatrix} a t_1^2 & 2 a t_1 & 1 \\ a\left(t_2^2-t_1^2\right) & 2 a\left(t_2-t_1\right) & 0 \\ a\left(t_3^2-t_2^2\right) & 2 a\left(t_3-t_2\right) & 0 \end{vmatrix}$
$ =\frac{1}{2}\begin{vmatrix} a\left(t_2^2-t_1^2\right) & 2 a\left(t_2-t_1\right) \\ a\left(t_3^2-t_2^2\right) & 2 a\left(t_3-t_2\right) \end{vmatrix}$
$ =\frac{1}{2} \cdot a \cdot 2 a , \begin{vmatrix} \left(t_2-t_1\right)\left(t_2+t_1\right) & \left(t_2-t_1\right) \\ \left(t_3-t_2\right)\left(t_3+t_2\right) & \left(t_3-t_2\right) \end{vmatrix}$
$=a^2\left(t_2-t_1\right)\left(t_3-t_2\right)$ $\begin{vmatrix} t_2+t_1 & 1 \\ t_3+t_2 & 1 \end{vmatrix} $
$=a^2\left|\left(t_2-t_1\right)\left(t_3-t_2\right)\left(t_1-t_3\right)\right|$
Again, let $\Delta_2=$ area of the $\triangle P Q R$
$ \begin{aligned} & =\frac{1}{2} \begin{vmatrix} a t_1 t_2 & a\left(t_1+t_2\right) & 1 \\ a t_2 t_3 & a\left(t_2+t_3\right) & 1 \\ a t_3 t_1 & a\left(t_3+t_1\right) & 1 \end{vmatrix} \\ & =\frac{1}{2} a \cdot a\begin{vmatrix} t_1 t_2 & \left(t_1+t_2\right) & 1 \\ t_2 t_3 & \left(t_2+t_3\right) & 1 \\ t_3 t_1 & \left(t_3+t_1\right) & 1 \end{vmatrix} \end{aligned} $
Applying $R_3 \rightarrow R_3-R_2, R_2 \rightarrow R_2-R_1$, we get
$ \begin{aligned} & =\frac{a^2}{2}\begin{vmatrix} t_1 t_2 & t_1+t_2 & 1 \\ t_2\left(t_3-t_1\right) & t_3-t_1 & 0 \\ t_3\left(t_1-t_2\right) & t_1-t_2 & 0 \end{vmatrix} \\ & =\frac{a^2}{2}\left(t_3-t_1\right)\left(t_1-t_2\right)\begin{vmatrix} t_1 t_2 & t_1+t_2 & 1 \\ t_2 & 1 & 0 \\ t_3 & 1 & 0 \end{vmatrix} \\ & =\frac{a^2}{2}\left(t_3-t_1\right)\left(t_1-t_2\right)\begin{vmatrix} t_2 & 1 \\ t_3 & 1 \end{vmatrix} \\ & =\frac{a^2}{2}\left|\left(t_3-t_1\right)\left(t_1-t_2\right)\left(t_2-t_3\right)\right| \end{aligned} $
Therefore, $\frac{\Delta_1}{\Delta_2}=\frac{a^2\left|\left(t_2-t_1\right)\left(t_3-t_2\right)\left(t_1-t_3\right)\right|}{\frac{1}{2} a^2\left|\left(t_3-t_1\right)\left(t_1-t_2\right)\left(t_2-t_3\right)\right|}=2$
BETA
