Question

Tangent and normal are drawn at $P\left( 16,16 \right)$ 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 $\angle CPB=\theta$, then a value of $\tan \theta$ is
(a) $3$
(b) $\dfrac{4}{3}$
(c) $\dfrac{1}{2}$
(d) $2$

Answer 1

By comparing the given equation of parabola with the standard equation ${{y}^{2}}=4ax$, we can find out the value of a. The parametric point on a parabola is given by $\left( a{{t}^{2}},2at \right)$ which on equating with the given coordinates $P\left( 16,16 \right)$ will give the value of t. The equation of tangent and normal at P will be obtained by substituting the value of t in $y=\dfrac{x}{t}+at$ and $y=-tx+2at+a{{t}^{3}}$ respectively. On substituting $y=0$ in these equations of tangent and normal, we will get the coordinates of A and B. The midpoint of A and B will give the coordinates of the centre C. Finally, using the formula $\tan \theta =\left| \dfrac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|$, we will get the final answer.

Complete step by step solution:
The equation of the parabola is given as
$\begin{aligned} & \Rightarrow {{y}^{2}}=16x \\\ & \Rightarrow {{y}^{2}}=4\left( 4 \right)x \\\ \end{aligned}$
Comparing the above equation with the standard equation of a parabola ${{y}^{2}}=4ax$, we get
$\Rightarrow a=4........\left( i \right)$
Now, the parametric point on a parabola is given as $\left( a{{t}^{2}},2at \right)$. On substituting $a=4$, we can write the parametric point as $\left( 4{{t}^{2}},8t \right)$.
The point P on the parabola is given as $\left( 16,16 \right)$. So we can equate the y-coordinate with the parametric y-coordinate $8t$ to get
$\begin{aligned} & \Rightarrow 8t=16 \\\ & \Rightarrow t=2........\left( ii \right) \\\ \end{aligned}$
Now, the equation for tangent in terms of the parameter t is given by
$\Rightarrow y=\dfrac{x}{t}+at$
Substituting (i) and (ii) in the above equation, we get
$\begin{aligned} & \Rightarrow y=\dfrac{x}{2}+4\left( 2 \right) \\\ & \Rightarrow y=\dfrac{x}{2}+8.......\left( iii \right) \\\ \end{aligned}$
Also, the equation for the normal in terms of parameter t is given by
$\Rightarrow y=-tx+2at+a{{t}^{3}}$
Substituting (i) and (ii) in the above equation, we get

& \Rightarrow y=-2x+2\left( 4 \right)\left( 2 \right)+4{{\left( 2 \right)}^{3}} \\\ & \Rightarrow y=-2x+16+32 \\\ & \Rightarrow y=-2x+48......\left( iv \right) \\\ \end{aligned}$$ According to the question, the tangent intersect the axis of the parabola, which is x-axis, at the point A. Therefore, we substitute $y=0$ in the equation (iii) to get the x-coordinate of A as $$\begin{aligned} & \Rightarrow 0=\dfrac{x}{2}+8 \\\ & \Rightarrow \dfrac{x}{2}+8=0 \\\ \end{aligned}$$ Subtracting $$8$$ from both sides, we get $\begin{aligned} & \Rightarrow \dfrac{x}{2}+8-8=0-8 \\\ & \Rightarrow \dfrac{x}{2}=-8 \\\ \end{aligned}$ Multiplying both sides by $2$ we get $\begin{aligned} & \Rightarrow \dfrac{x}{2}\times 2=-8\times 2 \\\ & \Rightarrow x=-16 \\\ \end{aligned}$ Therefore, the coordinates of A are $\left( -16,0 \right)$. Now, the normal intersects the axis at B. Therefore, we substitute $y=0$ in the equation (iv) to get the x-coordinate of B as: Subtracting $48$ from both sides, we get $\begin{aligned} & \Rightarrow -2x+48-48=0-48 \\\ & \Rightarrow -2x=-48 \\\ \end{aligned}$ Dividing both sides by $-2$ we get $\begin{aligned} & \Rightarrow \dfrac{-2x}{-2}=\dfrac{-48}{-2} \\\ & \Rightarrow x=24 \\\ \end{aligned}$ Therefore, the coordinates of B are $\left( 24,0 \right)$. Now, since the circle passes through the points P, A, and B, AB will be its diameter and the centre C will be the midpoint of AB. Since both the points A and B have zero y-coordinate, the y-coordinate of the centre C will also be equal to zero. And the x-coordinate of the centre will be given by the midpoint formula as $\begin{aligned} & \Rightarrow {{x}_{c}}=\dfrac{-16+24}{2} \\\ & \Rightarrow {{x}_{c}}=\dfrac{8}{2} \\\ & \Rightarrow {{x}_{c}}=4 \\\ \end{aligned}$ Therefore, the coordinates of the centre C are $\left( 4,0 \right)$. The circle is as shown in the below figure.  From the above figure, using the two point formula we can write the slope of the line CP as $$\begin{aligned} & \Rightarrow {{m}_{1}}=\dfrac{16-0}{16-4} \\\ & \Rightarrow {{m}_{1}}=\dfrac{16}{12} \\\ & \Rightarrow {{m}_{1}}=\dfrac{4}{3}.......\left( v \right) \\\ \end{aligned}$$ Similarly, we can write the slope of the line PB as $\begin{aligned} & \Rightarrow {{m}_{2}}=\dfrac{16-0}{16-24} \\\ & \Rightarrow {{m}_{2}}=\dfrac{16}{-8} \\\ & \Rightarrow {{m}_{2}}=-2.......\left( vi \right) \\\ \end{aligned}$ Now, since the angle $\theta $ lies between the lines CP and PB, the value of $\tan \theta $ can be given as $$\Rightarrow \tan \theta =\left| \dfrac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|$$ Substituting (v) and (vi) we get $$\begin{aligned} & \Rightarrow \tan \theta =\left| \dfrac{\dfrac{4}{3}-\left( -2 \right)}{1+\dfrac{4}{3}\left( -2 \right)} \right| \\\ & \Rightarrow \tan \theta =\left| \dfrac{\dfrac{4}{3}+2}{1-\dfrac{8}{3}} \right| \\\ & \Rightarrow \tan \theta =\left| \dfrac{\dfrac{10}{3}}{-\dfrac{5}{3}} \right| \\\ & \Rightarrow \tan \theta =\left| -2 \right| \\\ & \Rightarrow \tan \theta =2 \\\ \end{aligned}$$ Hence, the correct answer is option D. **Note:** We must remember the important results of the parabola in terms of the parameter t so as to quickly attempt these kinds of questions. We can also obtain the equations for the tangent and the normal by finding out their respective slopes and using the point slope form. But that would increase our calculations.