Question

Two parabolas have the same vertex and equal length of latus rectum such that their axes are at right angles. Prove that the common tangent touches each at the end of a latus rectum.

Answer 1

Hint:In this question, first work out on two parabolas whose vertices are the same and consider the equation of tangent in slope form for any one of the parabola. Second work out on the value of slope and finally put this value in the equation of the tangent.

Complete step-by-step answer:
Let the two parabolas be
${{y}^{2}}=4ax....................(1)$
and
${{x}^{2}}=4ay.....................(2)$
Equation of tangent to the parabola of equation (1) in slope form is given by
$y=mx+\dfrac{a}{m}$
It also touches to the parabola of equation (2), so substituting the value of y in equation (2), we get
${{x}^{2}}=4a\left( mx+\dfrac{a}{m} \right)$
Multiplying both sides by m, we get
$m{{x}^{2}}=4a{{m}^{2}}x+4{{a}^{2}}$
Rearranging the terms, we get
$m{{x}^{2}}-4a{{m}^{2}}x-4{{a}^{2}}=0$
This is a quadratic equation in x and it will have only one root.
So, discriminant = 0
${{B}^{2}}-4AC=0$
${{\left( -4a{{m}^{2}} \right)}^{2}}-4(m)(-4{{a}^{2}})=0$
$16{{a}^{2}}{{m}^{4}}+16{{a}^{2}}m=0$
Dividing both sides by $16{{a}^{2}}$ , we get
${{m}^{4}}+m=0$
Dividing both sides by m, we get
${{m}^{3}}+1=0$
${{m}^{3}}=-1={{(-1)}^{3}}$
$m=-1$
For point of contact (Parabola ${{y}^{2}}=4ax$and a line\[y=mx+c\])
We solve the two equations ${{y}^{2}}=4ax$and \[y=mx+c\] simultaneously, we get
$\begin{align}
  & {{y}^{2}}=4a\left( \dfrac{y-c}{m} \right) \\
 & m{{y}^{2}}=4ay-4ac \\
 & m{{y}^{2}}-4ay+4ac=0 \\
\end{align}$
Now put $c=\dfrac{a}{m}$ in the above equation, we get
$\begin{align}
  & m{{y}^{2}}-4ay+4a\left( \dfrac{a}{m} \right)=0 \\
 & {{m}^{2}}{{y}^{2}}-4aym+4{{a}^{2}}=0 \\
 & {{\left( my-2a \right)}^{2}}=0 \\
 & my-2a=0 \\
 & y=\dfrac{2a}{m} \\
\end{align}$
Putting $y=\dfrac{2a}{m}$ either in the equation of parabola ${{y}^{2}}=4ax$, we get $x=\dfrac{a}{{{m}^{2}}}$
The point of contact to parabola which is represented by equation (1) is $\left( \dfrac{a}{{{m}^{2}}},\dfrac{2a}{m} \right)$
Therefore, $\left( a,-2a \right)$
For point of contact (Parabola ${{x}^{2}}=4ay$ and a line\[x=my+c\])
We solve the two equations ${{x}^{2}}=4ay$and \[x=my+c\] simultaneously, we get
$\begin{align}
  & {{x}^{2}}=4a\left( \dfrac{x-c}{m} \right) \\
 & m{{x}^{2}}=4ax-4ac \\
 & m{{x}^{2}}-4ax+4ac=0 \\
\end{align}$
Now put $c=\dfrac{a}{m}$ in the above equation, we get
$\begin{align}
  & m{{x}^{2}}-4ax+4a\left( \dfrac{a}{m} \right)=0 \\
 & {{m}^{2}}{{x}^{2}}-4axm+4{{a}^{2}}=0 \\
 & {{\left( mx-2a \right)}^{2}}=0 \\
 & mx-2a=0 \\
 & x=\dfrac{2a}{m} \\
\end{align}$
Putting $x=\dfrac{2a}{m}$ either in the equation of parabola${{x}^{2}}=4ay$, we get $y=\dfrac{a}{{{m}^{2}}}$
The point of contact to parabola which is represented by equation (2) is $\left( \dfrac{2a}{m},\dfrac{a}{{{m}^{2}}} \right)$
Therefore, $\left( -2a,a \right)$
Hence, $\left( a,-2a \right)$and $\left( -2a,a \right)$ are the endpoints of the parabolas ${{y}^{2}}=4ax$ and ${{x}^{2}}=4ay$ respectively.
Therefore, the common tangent touches each at the end points of a latus rectum.

Note: The parabolas are ${{y}^{2}}=4ax$ and ${{x}^{2}}=4ay$. The tangent of the first parabola at a point (a, -2a) that is at the one end of the latus rectum is y = x + a and put value of y at the other equation you get, it cuts the other parabola at the point (-2a, a) that is at the end of the latus rectum (You should note that we have used the direct formulae of the tangent equation $y{{y}_{1}}=2a(x+{{x}_{1}})$ ).