Question

Prove that two parabolas, having the same focus and their axes in opposite directions, cut at right angles.

Answer 1

Hint: We will assume two different parabolas having equations ${{y}^{2}}=4ax$ and ${{y}^{2}}=-4a\left( x-2a \right)$. We will find the intersecting points of the parabolas and then by differentiating the equation of parabola, we will find the slope of tangents. It is noted that if the product of tangents is equal to (-1), then, the tangents are perpendicular to each other.

Complete step by step solution:
It is given in the question that we have to prove that two parabolas, having the same focus and their axes in opposite directions, cut at right angles.
So, let us assume that the equation of the first parabola will be, ${{y}^{2}}=4ax.........\left( i \right)$
And let us assume that the equation of the second parabola will be ${{y}^{2}}=-4a\left( x-2a \right).........\left( ii \right)$
Now, on comparing equations (i) and (ii), we get,
$4ax=-4a\left( x-2a \right)$
On cancelling the like terms on both sides, we get,
$x=-x+2a$
On transposing (-x) from the RHS to LHS, we get,
$\begin{aligned} & x+x=2a \\\ & 2x=2a \\\ \end{aligned}$
On dividing the above expression by 2, we get,
$\begin{aligned} & \dfrac{2x}{2}=\dfrac{2a}{2} \\\ & x=a \\\ \end{aligned}$
Now, on substituting the value of x = a in equation (i), we get,
$\begin{aligned} & {{y}^{2}}=4a\left( a \right) \\\ & {{y}^{2}}=4{{a}^{2}} \\\ \end{aligned}$
On taking the square root on both the sides, we get,
$\begin{aligned} & \sqrt{{{y}^{2}}}=\sqrt{4{{a}^{2}}} \\\ & y=\pm 2a \\\ \end{aligned}$
Now, if we substitute x = a in equation (ii), we get,
$\begin{aligned} & {{y}^{2}}=-4a\left( a-2a \right) \\\ & {{y}^{2}}=-4a\left( -a \right) \\\ & {{y}^{2}}=4{{a}^{2}} \\\ \end{aligned}$
On taking the square root on both the sides, we get,
$\begin{aligned} & \sqrt{{{y}^{2}}}=\sqrt{4{{a}^{2}}} \\\ & y=\pm 2a \\\ \end{aligned}$
It means that the parabolas are intersecting at the point $\left( a,\pm 2a \right)$.
We know that slope of a tangent is given by differentiating the equation of the parabola. So, slope of tangents to parabola (i) at the point of intersection is,
$\begin{aligned} & \dfrac{d\left( {{y}^{2}} \right)}{dx}=\dfrac{d\left( 4ax \right)}{dx} \\\ & 2y\dfrac{dy}{dx}=4a \\\ & \dfrac{dy}{dx}=\dfrac{4a}{2y} \\\ & \dfrac{dy}{dx}=\dfrac{2a}{y} \\\ \end{aligned}$
On putting $y=\pm 2a$, we get,
$\dfrac{dy}{dx}=\dfrac{2a}{\pm 2a}$
Now, we know that $\dfrac{dy}{dx}$ gives us the value of slope. So, we can write the slope of the first parabola, $\dfrac{dy}{dx}$ as ${{m}_{1}}$. So, we can write,
${{m}_{1}}=\pm 1$
Now, we will find the slope of tangent to parabola (ii) at the point of intersection.

& \dfrac{d\left( {{y}^{2}} \right)}{dx}=\dfrac{d\left( -4a\left( x-2a \right) \right)}{dx} \\\ & \dfrac{d\left( {{y}^{2}} \right)}{dx}=\dfrac{d\left( -4ax+8{{a}^{2}} \right)}{dx} \\\ & 2y\dfrac{dy}{dx}=-4a \\\ & \dfrac{dy}{dx}=\dfrac{-4a}{2y} \\\ & \dfrac{dy}{dx}=\dfrac{-2a}{y} \\\ \end{aligned}$$ On substituting $y=\pm 2a$, we get, $\begin{aligned} & \dfrac{dy}{dx}=\dfrac{-2a}{\pm 2a} \\\ & \dfrac{dy}{dx}=\dfrac{2a}{\mp 2a} \\\ \end{aligned}$ Now, we know that $\dfrac{dy}{dx}$ gives us the value of slope. So, we can write the slope of the second parabola, $\dfrac{dy}{dx}$ as ${{m}_{2}}$. So, we can write, ${{m}_{2}}=\mp 1$ Now, we know that if the product of two slopes or tangents is equal to (-1), then the tangents are perpendicular to each other. So, here we get, $\begin{aligned} & {{m}_{1}}{{m}_{2}}=\left( \pm 1 \right)\left( \mp 1 \right) \\\ & {{m}_{1}}{{m}_{2}}=-1 \\\ \end{aligned}$ Hence, we have proved that two parabolas, having the same focus and their axes in opposite directions, cut at right angles. Note: While solving this question, many students get confused while finding ${{m}_{2}}$, and may skip the minus sign before 2a and due to this they will get, ${{m}_{2}}=\pm 1$. So, in the last step, while taking the product of the two slopes, they we get, ${{m}_{1}}{{m}_{2}}=\left( \pm 1 \right)\left( \pm 1 \right)=1$ and not equal to (-1). Therefore, the students must be careful while solving this question.