Parabola Ques 47
- 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)
Show Answer
Answer:
Correct Answer: 47.$(-1,0)$
Solution:
- 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 $2 y \rightarrow y+y _1$ ]
$$ \begin{alignedat} & \Rightarrow \quad 2x = y - 5 \` & \Rightarrow \quad y=2 x+5 \end{aligned} $$
which is also tangent to the circle’s circumference
$$ x^{2}+y^{2}+16 x+12 y+c=0 $$
i.e. $x^{2}+(2 x+5)^{2}+16 x+12(2 x+5)+C=0$ must have equal roots i.e., $\alpha=\beta$
$$ \begin{alignedat} & \Rightarrow \quad 5 x^{2}+60 x+85+c=0 \\ & \Rightarrow \quad \alpha+\beta=\frac{-60}{5} \\ & \Rightarrow \quad \alpha=-6 \\ & \therefore \quad x=-6 \text { and } y=2 x+5=-7 \end{aligned} $$
$\therefore$ Point of contact is $(-6,-7)$.
BETA
