SATHEE: Parabola 3 Question 9

Parabola 3 Question 9

9. 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 $m_{1} m_{2} = α$ is a part of the parabola itself, then find $α$ .

(2003, 4M)

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Answer:

Correct Answer: 9. (2)

Solution:

Let the three points on the parabola be $A (a t_{1}^{2}, 2 a t_{1}), B (α t_{2}^{2}, 2 a t_{2})$ and $C (a t_{3}^{2}, 2 a t_{3})$ .

Equation of the tangent to the parabola at $(a t^{2}, 2 a t)$ is

$t y = x + a t^{2}$

Therefore, equations of tangents at $A$ and $B$ are

$\begin{aligned} t_{1} y = x + a t_{1}^{2} \\ t_{2} y = x + a t_{2}^{2} \end{aligned}$

From Eqs. (i) and (ii)

$\begin{aligned} t_{1} y = t_{2} y - a t_{2}^{2} + a t_{1}^{2} \\ \Rightarrow t_{1} y - t_{2} y = a t_{1}^{2} - a t_{2}^{2} \\ \Rightarrow y = a (t_{1} + t_{2}) \\ and t_{1} a (t_{1} + t_{2}) = x + a t_{1}^{2} \\ \Rightarrow x = a t_{1} t_{2} \end{aligned}$

$[∵ t_{1} \neq t_{2}]$

Therefore, coordinates of $P$ are $(a t_{1} t_{2}, a (t_{1} + t_{2}))$

Similarly, the coordinates of $Q$ and $R$ are respectively,

$[a t_{2} t_{3}, a (t_{2} + t_{3})] and [a t_{1} t_{3}, a (t_{1} + t_{3})]$

Let $Δ_{1} =$ Area of the $△ A B C$

$= \frac{1}{2} | \begin{array}{lll} a t_{1}^{2} & 2 a t_{1} & 1 \\ a t_{2}^{2} & 2 a t_{2} & 1 \\ a t_{3}^{2} & 2 a t_{3} & 1 \end{array} | |$

Applying $R_{3} \to R_{3} - R_{2}$ and $R_{2} \to R_{2} - R_{1}$ , we get

$\begin{aligned} Δ_{1} = \frac{1}{2} | \begin{array}{ccc} a t_{1}^{2} & 2 a t_{1} & 1 \\ a (t_{2}^{2} - t_{1}^{2}) & 2 a (t_{2} - t_{1}) & 0 \\ a (t_{3}^{2} - t_{2}^{2}) & 2 a (t_{3} - t_{2}) & 0 \end{array} | \\ = \frac{1}{2} | | \begin{array}{ccc} a (t_{2}^{2} - t_{1}^{2}) & 2 a (t_{2} - t_{1}) \\ a (t_{3}^{2} - t_{2}^{2}) & 2 a (t_{3} - t_{2}) \end{array} | | \\ = \frac{1}{2} \cdot a \cdot 2 a | \begin{array}{ccc} (t_{2} - t_{1}) & (t_{2} + t_{1}) & (t_{2} - t_{1}) \\ (t_{3} - t_{2}) & (t_{3} + t_{2}) & (t_{3} - t_{2}) \end{array} ∣ \\ = a^{2} (t_{2} - t_{1}) (t_{3} - t_{2}) | | \begin{array}{cc} t_{2} + t_{1} & 1 \\ t_{3} + t_{2} & 1 \end{array} | | \\ = a^{2} | (t_{2} - t_{1}) (t_{3} - t_{2}) (t_{1} - t_{3}) | \end{aligned}$

Again, let $Δ_{2} =$ area of the $△ P Q R$

$\begin{aligned} = \frac{1}{2} | | \begin{array}{lll} a t_{1} t_{2} & a (t_{1} + t_{2}) & 1 \\ a t_{2} t_{3} & a (t_{2} + t_{3}) & 1 \\ a t_{3} t_{1} & a (t_{3} + t_{1}) & 1 \end{array} | | \\ = \frac{1}{2} a \cdot a | | \begin{array}{lll} t_{1} t_{2} & (t_{1} + t_{2}) & 1 \\ t_{2} t_{3} & (t_{2} + t_{3}) & 1 \\ t_{3} t_{1} & (t_{3} + t_{1}) & 1 \end{array} | | \end{aligned}$

Applying $R_{3} \to R_{3} - R_{2}, R_{2} \to R_{2} - R_{1}$ , we get

$\begin{aligned} = \frac{a^{2}}{2} | | \begin{array}{ccc} t_{1} t_{2} & t_{1} + t_{2} & 1 \\ t_{2} (t_{3} - t_{1}) & t_{3} - t_{1} & 0 \\ t_{3} (t_{1} - t_{2}) & t_{1} - t_{2} & 0 \end{array} | | \\ = \frac{a^{2}}{2} (t_{3} - t_{1}) (t_{1} - t_{2}) | \begin{array}{ccc} t_{1} t_{2} & t_{1} + t_{2} & 1 \\ t_{2} & 1 & 0 \\ t_{3} & 1 & 0 \end{array} | ∣ \\ = \frac{a^{2}}{2} (t_{3} - t_{1}) (t_{1} - t_{2}) | \begin{array}{cc} t_{2} & 1 \\ t_{3} & 1 \end{array} | \\ = \frac{a^{2}}{2} | (t_{3} - t_{1}) (t_{1} - t_{2}) (t_{2} - t_{3}) | \end{aligned}$

Therefore, $\frac{Δ_{1}}{Δ_{2}} = \frac{a^{2} | (t_{2} - t_{1}) (t_{3} - t_{2}) (t_{1} - t_{3}) |}{\frac{1}{2} a^{2} | (t_{3} - t_{1}) (t_{1} - t_{2}) (t_{2} - t_{3}) |} = 2$

Parabola 3 Question 9

9. 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 $m_{1} m_{2} = α$ is a part of the parabola itself, then find $α$ .

Answer:

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Parabola 3 Question 9

9. Normals are drawn from the point P with slopes m1,m2,m3 to the parabola y2=4x. If locus of P with m1m2=α is a part of the parabola itself, then find α.

Answer:

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9. 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 $m_{1} m_{2} = α$ is a part of the parabola itself, then find $α$ .