Question

If the normal to ${{y}^{2}}=12x$ at $\left( 3,6 \right)$ meets the parabola again in $\left( 27,-18 \right)$ and the circle on the normal chord as the diameter is
(a) ${{x}^{2}}+{{y}^{2}}+30x+12y-27=0$
(b) ${{x}^{2}}+{{y}^{2}}+30x+12y+27=0$
(c) ${{x}^{2}}+{{y}^{2}}-30x-12y-27=0$
(d) ${{x}^{2}}+{{y}^{2}}-30x+12y-27=0$

Answer 1

Hint: We have to confirm the endpoints of the normal chord, for which we can use the parametric form of a point lying on the parabola $\left( a{{t}^{2}},2at \right)$ and also the equation of the normal $y+tx=2at+a{{t}^{3}}$.

Complete step-by-step solution -
Let us consider the parabola given in the question, ${{y}^{2}}=12x$. On comparing it with the general equation of the parabola ${{y}^{2}}=4ax$ we get,

& 4a=12 \\\ & a=3 \\\ \end{aligned}$$ It is given in the question that the point $$\left( 3,6 \right)$$ lies on the parabola. We know that the parametric form of a point lying on the parabola is given by $$\left( a{{t}^{2}},2at \right)$$. So, on comparing the $$x$$ coordinate, we get $$\begin{aligned} & a{{t}^{2}}=3 \\\ & \therefore a=3 \\\ & 3\times {{t}^{2}}=3 \\\ & {{t}^{2}}=1 \\\ \end{aligned}$$ Now, on comparing the $y$ coordinate, we get $$\begin{aligned} & 2at=6 \\\ & \therefore a=3 \\\ & 2\times 3\times t=6 \\\ & 6t=6 \\\ & t=1 \\\ \end{aligned}$$ Therefore, we get the value as $$t=1$$. Now, we know that the equation of the normal is given by $y+tx=2at+a{{t}^{3}}$. Substituting the values of $$a=3$$ and $$t=1$$ in it, we get $\begin{aligned} & y+x=2\times 3\times 1+3\times {{1}^{3}} \\\ & y+x=6+3 \\\ & y+x=9 \\\ & y=9-x \\\ \end{aligned}$ It is known that this normal intersects the parabola ${{y}^{2}}=12x$, so they must satisfy the equations of each other. Substituting $$y=9-x$$ in ${{y}^{2}}=12x$, we get ${{\left( 9-x \right)}^{2}}=12x$ Since we know that $${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$$, we get $$\begin{aligned} & 81-18x+{{x}^{2}}=12x \\\ & {{x}^{2}}-30x+81=0 \\\ & {{x}^{2}}-27x-3x+81=0 \\\ & x\left( x-27 \right)-3\left( x-27 \right)=0 \\\ & \left( x-3 \right)\left( x-27 \right)=0 \\\ \end{aligned}$$ Therefore, we get the value as $$x=3$$ or $$x=27$$. Now, from the question, it is clear that the point $$\left( 3,6 \right)$$ already lies on the normal. So, using $$x=27$$to find the other point, we get $\begin{aligned} & y=9-x \\\ & y=9-27 \\\ & y=-18 \\\ \end{aligned}$ So, we get the point as $\left( 27,-18 \right)$. Since the circle is formed on the diameter which is the normal chords, the ends of the diameter are $$\left( 3,6 \right)$$ and $\left( 27,-18 \right)$. Thus, by using the midpoint formula given by $\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)$, we can get the center of the circle as, $$\begin{aligned} & \left( \dfrac{3+27}{2},\dfrac{6-18}{2} \right) \\\ & \left( \dfrac{30}{2},\dfrac{-12}{2} \right) \\\ & \left( 15,-6 \right) \\\ \end{aligned}$$ Now, we can plot a graph and consider the figure below. We can find the radius by using the distance formula given by $$\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$$. Referring the figure for coordinates, we get the radius as, $$\begin{aligned} & \sqrt{{{\left( 15-3 \right)}^{2}}+{{\left( -6-6 \right)}^{2}}} \\\ & \sqrt{{{\left( 12 \right)}^{2}}+{{\left( -12 \right)}^{2}}} \\\ & \sqrt{144+144} \\\ & \sqrt{288} \\\ \end{aligned}$$ Thus the equation of the circle can be obtained using the general form ${{\left( x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}={{r}^{2}}$. We have the coordinates of centre $$\left( a,b \right)=\left( 15,-6 \right)$$ and radius $$r=\sqrt{288}$$. So, the equation of the circle can be formulated as, ${{\left( x-15 \right)}^{2}}+{{\left( y+6 \right)}^{2}}={{\left( \sqrt{288} \right)}^{2}}$ Since we know that $${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$$ and also $${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$$, we get $\begin{aligned} & {{x}^{2}}-30x+225+{{y}^{2}}+12y+36=288 \\\ & {{x}^{2}}+{{y}^{2}}-30x+12y+261=288 \\\ & {{x}^{2}}+{{y}^{2}}-30x+12y-27=0 \\\ \end{aligned}$ Therefore, we get the circle as ${{x}^{2}}+{{y}^{2}}-30x+12y-27=0$ and option (d) is the correct answer. Note: An alternative approach can be used by substituting the points $$\left( 3,6 \right)$$ and $\left( 27,-18 \right)$, one by one in all the given options because if they are the ends of the diameter they must satisfy the equation of the circle and there is only a unique circle ${{x}^{2}}+{{y}^{2}}-30x+12y-27=0$ satisfying both the points.