Parabola Ques 58
- Normals are drawn from the point $P$ with slopes $m _1, m _2, m _3$ to the parabola $y^{2}=4 x$. If locus of $P$ with $m _1 m _2=\alpha$ is a part of the parabola itself, then find $\alpha$.
(2003, 4M)
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Answer:
Correct Answer: 58.(2)
Solution:
Formula:
- Let the three points on the parabola be $A\left(a t _1^{2}, 2 a t _1\right), B\left(\alpha t _2^{2}, 2 a t _2\right)$ 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 points $A$ and $B$ are
$$ \begin{aligned} & t_1 y=x+a t_1^{2} \\ & t_2 y=x+a t_2^{2} \end{aligned} $$
From Eqs. (i) and (ii)
$$ \begin{aligned} & t _1 y=t _2 y-a t _2^{2}+a t _1^{2} \\ & \Rightarrow \quad t _1 y-t _2 y=a t _1^{2}-a t _2^{2} \\ & \Rightarrow \quad y=a\left(t _1+t _2\right) \\ & \text { and } \quad t _1 a\left(t _1+t _2\right)=x+a t _2^{2} \\ & \Rightarrow \quad x= a t_1 t_2 \end{aligned} $$
$$ \left[\because t _1 \neq t _2\right] $$
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,
$$ \left[a t _2 t _3, a\left(t _2+t _3\right)\right] \text { and }\left[a t _1 t _3, a\left(t _1+t _3\right)\right] $$
Let $\quad \Delta _1=$ Area of $\triangle ABC$
$$ =$\frac{1}{2} | \begin{array}{ccc} a t _1^{2} & 2 a t _1 & 1 \ a t _{2}^{2} & 2 a t _{2} & 1 \\ a t _3^{2} & 2 a t _3 & 1 \end{array} $$
Applying $R _3 \rightarrow R _3-R _2$ and $R _2 \rightarrow R _2-R _1$, we get
$$ \begin{aligned} & \Delta _1=\frac{1}{2}\left|\begin{array}{ccc} 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{array}\right| \\ & =\frac{1}{2}|| \begin{array}{ccc} 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{array} \ \\ & =\frac{1}{2} \cdot a \cdot 2 a | \begin{array}{ccc} \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{array} \mid \\ & =a^{2}\left(t{2}-t{1}\right)\left(t{3}-t{2}\right)|| \begin{array}{cc} t{2}+t{1} & 1 \) t_3 + t_2 & 1 \end{array} \ \\ & =a^{2}\left|\left(t _2-t _1\right)\left(t _3-t _2\right)\left(t _1-t _3\right)\right| \end{aligned} $$
Again, let $\Delta _2=$ area of the $\triangle P Q R$
$$ \begin{aligned} & =\frac{1}{2}|| \begin{array}{lll} 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{array} \ \\ & =\frac{1}{2} a \cdot a|| \begin{array}{lll} 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{array} \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{array}{c} 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{array} \ \\ & =\frac{a^{2}}{2}\left(t_3-t_1\right)\left(t_1-t_2\right)\left|\begin{array}{ccc} t{1} t{2} & t{1}+t{2} & 1 \) t _2 & 1 & 0 \ t _3 & 1 & 0 \end{array}\right| \mid \\ & =\frac{a^{2}}{2}\left(t_3-t_1\right)\left(t_1-t_2\right)\left|\begin{array}{cc} t _2 & 1 \ t _3 & 1 \end{array}\right| \\ & =\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
