Tangent and normal are drawn at P(16,16) on the parabola y^2=16x, which intersect the axis…

Question

Tangent and normal are drawn at $P(16,16)$ on the parabola $y^2=16x$, which intersect the axis of the parabola at $A$ and $B$, respectively. If $C$ is the centre of the circle through the points $P$, $A$ and $B$ and $\angle CPB=	heta$, then a value of $	an	heta$ is:

Accepted Answer

Tangent at $P$: $16y=8(x+16)\Rightarrow y=\dfrac x2+8$; at $y=0$: $A=(-16,0)$.

Normal at $P$ (slope $-2$): $y=-2x+48$; at $y=0$: $B=(24,0)$.

Centre $C=(4,k)$ (on perpendicular bisector of $AB$, $x=4$); equating distances to $A$ and $P$ gives $k=0$, so $C=(4,0)$.

Vectors from $P$: $\vec{PC}=(-12,-16)$, $\vec{PB}=(8,-16)$.

$$\cos\theta=\frac{\vec{PC}\cdot\vec{PB}}{|\vec{PC}||\vec{PB}|}=\frac{160}{20\sqrt{320}}=\frac1{\sqrt5},\qquad \sin\theta=\frac{320}{20\sqrt{320}}=\frac2{\sqrt5}.$$

$$\tan\theta=2.$$

Solutions · 1