Parabola Ques 36
- The point of intersection of the tangents at the ends of the latusrectum of the parabola $y^{2}=4 x$ is … .
(1994, 2M)
Analytical & Descriptive Questions
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Solution:
- The coordinates of extremities of the latusrectum of $y^{2}=4 x$ are $(1,2)$ and $(1,-2)$.
Equations of tangents at these points are
$$ \begin{array}{rlrl} & y \cdot 2=\frac{4(x+1)}{2} \Rightarrow 2 y & =2(x+1) \\ \text { and } & & y(-2) & =\frac{4(x+1)}{2} \\ \Rightarrow & & -2 y & =2(x+1) \end{array} $$
The point of intersection of these tangents can be obtained by solving Eqs. (i) and (ii) simultaneously.
$$ \begin{aligned} \therefore & & -2(x+1) & =2(x+1) \\ \Rightarrow & & 0 & =4(x+1) \\ \Rightarrow & & -1 & =x \Rightarrow y=0 \end{aligned} $$
Therefore, the required point is $(-1,0)$.
BETA
