Question

Let P be the point of intersection of the common tangents to the parabola ${{y}^{2}}=12x$ and the hyperbola $8{{x}^{2}}-{{y}^{2}}=8$ . If S and S’ denote the foci of the hyperbola where S lies on the positive x-axis then P divides SS' in a ratio ?\[\]
A.$5:4$ \[\]
B. $14:13$ \[\]
C. $2:1$\[\]
D. $13:11$\[\]

Answer 1

Hint: Compare the equation of parabola with general equation ${{y}^{2}}=4ax$to find $a$use it in the equation of tangent to the parabola at any point $y=mx+\dfrac{a}{m}$. Compare the equation of hyperbola with general equation $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ to find ${{a}^{2}},{{b}^{2}}$ and use it in the pair of equations of tangents to the hyperbola at any point $y=mx\pm \sqrt{{{a}^{2}}{{m}^{2}}-{{b}^{2}}}$. Equate respective sides and obtain a quadratic equation whose roots are possible values of $m$. Use that to obtain eccentricity and thereafter foci $S\left( ae,0 \right){{S}^{'}}\left( -ae,0 \right)$ . Use the section formula to find the ratio at which P divides $S{{S}^{'}}$.

Complete step-by-step answer:
The given equations are
\[\begin{align}
  & {{y}^{2}}=12x...\left( 1 \right) \\
 & 8{{x}^{2}}-{{y}^{2}}=8.....\left( 2 \right) \\
\end{align}\]

We can see equation(1) is an equation of parabola and equation(2) is an equation of hyperbola.
We know from differential calculus that the slope of any curve at any point is given by the differentiation with respect to the independent variable. We also know that the slope of the curve at any point is the slope of the tangent at that point. As curves (1) and (2) have common tangents we assign the possible values of slope as $m$
The general equation of any parabola is ${{y}^{2}}=4ax$ and the equation of tangent at any point on the parabola is given by
\[y=mx+\dfrac{a}{m}\]
We deduce from the equation(1) that \[{{y}^{2}}=4\cdot 3x\Rightarrow a=4\]. So the equation of the tangent transforms to
\[y=mx+\dfrac{3}{m}...(3)\]
The general equation of any hyperbola is $\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ and the equation of pair of tangents at any point on the hyperbola is given by
\[y=mx\pm \sqrt{{{a}^{2}}{{m}^{2}}-{{b}^{2}}}\]
We deduce from the equation(1) that \[8{{x}^{2}}-{{y}^{2}}=1\Rightarrow {{a}^{2}}=\dfrac{1}{8},{{b}^{2}}=1\]. So the equation of the tangent transforms to
\[y=mx\pm \sqrt{{{m}^{2}}-8}..\left( 4 \right)\]
We compare equation (3) and (4) are the same line as they are the common tangents. Let us equate right hand sides of equation (3) and (4)
\[\begin{align}
  & \dfrac{3}{m}=\pm \sqrt{{{m}^{2}}-8} \\
 & \Rightarrow 9={{m}^{2}}\left( {{m}^{2}}-8 \right) \\
\end{align}\]
Let us put ${{t}^{2}}=m$ in the above equation and solve the resultant quadratic equation,
\[\begin{align}
  & t\left( t-8 \right)=9 \\
 & \Rightarrow {{t}^{2}}-8t-9=0 \\
 & \Rightarrow t=-1,9 \\
 & \Rightarrow {{m}^{^{2}}}=9\left( \because {{m}^{2}}\ge 0 \right) \\
 & \Rightarrow m=3,-3 \\
\end{align}\]
Putting these values in equation(3) we equations of tangents ,
\[\begin{align}
  & y=3x+1 \\
 & y=-3x-1 \\
\end{align}\]
Solving the above pair of equations we get the point of intersection at $P\left( \dfrac{-1}{3},0 \right)$ . We use the relation ${{b}^{2}}={{a}^{2}}\left( {{e}^{2}}-1 \right)$ to find out eccentricity. Here $e=\pm 3$. The co-ordinates foci of the hyperbola are $S\left( ae,0 \right){{S}^{'}}\left( -ae,0 \right)=S\left( 3,0 \right),{{S}^{'}}\left( -3,0 \right)$ .
Let us use section formula for internal division and the point $P\left( \dfrac{k{{x}_{2}}+{{x}_{1}}}{k+1},\dfrac{k{{y}_{2}}+{{y}_{1}}}{k+1} \right)$ divide $S{{S}^{'}}$ at ratio $k:1$. So
\[\begin{align}
  & \dfrac{-1}{3}=\dfrac{-3k+3}{k+1} \\
 & \Rightarrow k=5:4 \\
\end{align}\]

So, the correct answer is “Option A”.

Note: We need to take care of the fact that unlike ellipse whose eccentricity lies between 0 to 1 the eccentricity of hyperbola is greater than 1. We have rejected the negative value of eccentricity because eccentricity is always positive. We also need to take care of the confusion between section formula with internal division from external division whose formula is $P\left( \dfrac{k{{x}_{2}}-{{x}_{1}}}{k-1},\dfrac{k{{y}_{2}}-{{y}_{1}}}{k-1} \right).$