Parabola 2 Question 30

9. The tangent at $(1,7)$ to the curves $x^{2}=y-6 x$ touches the circle $x^{2}+y^{2}+16 x+12 y+c=0$ at

(a) $(6,7)$

(b) $(-6,7)$

(c) $(6,-7)$

(d) $(-6,-7)$

(2005 2M)

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Answer:

Correct Answer: 9. $(-1,0)$

Solution:

  1. The tangent at $(1,7)$ to the parabola $x^{2}=y-6 x$ is

$$ x(1)=\frac{1}{2}(y+7)-6 $$

[replacing $x^{2} \rightarrow x x _1$ and $2 y \rightarrow y+y _1$ ]

$$ \begin{aligned} & \Rightarrow \quad 2 x=y+7-12 \\ & \Rightarrow \quad y=2 x+5 \end{aligned} $$

which is also tangent to the circle

$$ x^{2}+y^{2}+16 x+12 y+c=0 $$

i.e. $x^{2}+(2 x+5)^{2}+16 x+12(2 x+5)+C=0$ must have equal rools i.e., $\alpha=\beta$

$$ \begin{aligned} & \Rightarrow \quad 5 x^{2}+60 x+85+c=0 \\ & \Rightarrow \quad \alpha+\beta=\frac{-60}{5} \\ & \Rightarrow \quad \alpha=-6 \\ & \therefore \quad x=-6 \text { and } y=2 x+5=-7 \end{aligned} $$

$\therefore$ Point of contact is $(-6,-7)$.



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