Parabola Ques 21
- Let $P$ be the point on the parabola, $y^{2}=8 x$, which is at a minimum distance from the centre $C$ of the circle, $x^{2}+(y+6)^{2}=1$. Then, the equation of the circle, passing through $C$ and having its centre at $P$ is
(2016 Main)
(a) $x^{2}+y^{2}-4 x+8 y+12=0$
(b) $x^{2}+y^{2}-x+4 y-12=0$
(c) $x^{2}+y^{2}-\frac{x}{4}+2 y-24=0$
(d) $x^{2}+y^{2}-4 x+9 y+18=0$
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Answer:
Correct Answer: 21.(c)
Solution:
- Centre of circle $x^{2}+(y+6)^{2}=1$ is $C(0,-6)$.
Let the coordinates of point $P$ be $\left(2 t^{2}, 4 t\right)$.
Now, let’s
$$ \begin{aligned} D & =C P \ & =\sqrt{\left(2 t^{2}\right)^{2}+(4 t+6)^{2}} \end{aligned} $$
$$ \Rightarrow $$
Squaring on both sides
$$ \begin{array}{lrl} \Rightarrow & D^{2}(t) & =4 t^{4}+16 t^{2}+48 t+36 \\ \text { Let } & F(t) & =4 t^{4}+16 t^{2}+48 t+36 \\ \text { For minimum, } \quad F^{\prime}(t) & =0 \\ \Rightarrow \quad 16 t^{3}+32 t+48 & =0 \\ \Rightarrow \quad & t^{3}+2 t+3 & =0 \\ \Rightarrow \quad & (t+1)\left(t^{2}-t+3\right) & =0 \Rightarrow t=-1 \text{ or } t=\frac{1\pm\sqrt{-11}}{2} \end{array} $$
Thus, coordinate of point $P$ are $(2,-4)$.
Now.
$$ C P=\sqrt{2^{2}+(-4+6)^{2}}=\sqrt{4+4}=2 \sqrt{2} $$
Hence, the required equation of circle is
$$ \begin{aligned} & & (x-2)^{2}+(y+4)^{2} & =(2 \sqrt{2})^{2} \\ & \Rightarrow & x^{2}+4-4 x+y^{2}+16+8 y & =8 \\ \Rightarrow & & x^{2}+y^{2}-4 x+8 y+12 & =0 \end{aligned} $$
BETA
