Parabola 3 Question 6
6. Let $L$ be a normal to the parabola $y^{2}=4 x$. If $L$ passes through the point $(9,6)$, then $L$ is given by
(2011)
(a) $y-x+3=0$
(b) $y+3 x-33=0$
(c) $y+x-15=0$
(d) $y-2 x+12=0$
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Answer:
Correct Answer: 6. $(a, b, d)$
Solution:
- $m _{O P}=\frac{2 a t-0}{a t^{2}-0}=\frac{2}{t}$
$$ m _{O Q}=\frac{-2 a / t}{a / t^{2}}=-2 t $$
$\therefore \quad \tan \theta=\frac{\frac{2}{t}+2 t}{1-\frac{2}{t} \cdot 2 t}=\frac{2 t+\frac{1}{t}}{1-4}=\frac{-2 \sqrt{5}}{3}$
where $\quad t+\frac{1}{t}=\sqrt{5}$
