Question

Two parabolas ${{y}^{2}}=4a(x-{{m}_{1}})$ and ${{x}^{2}}=4a(y-{{m}_{2}})$ always touch one another, the quantities ${{m}_{1}}$, and ${{m}_{2}}$ are both variables. The locus of this point of contact has the equation
$(a)$ $xy = {a}^{2}$
$(b)$ $xy = {2{a}^{2}}$
$(c)$ $xy= {4{a}^{2}}$
$(d)$ $\text{None}$

Answer 1

Hint: Use the fact that the slope of tangents of both the parabolas is equal at the point at which they touch each other. Calculate the first derivative of the two equations of parabolas to find the slope of parabolas at the point at which they touch each other. Solve them to get the locus of this point of contact of the two parabolas.

Complete step-by-step solution -
We have two parabolas whose equations are ${{y}^{2}}=4a(x-{{m}_{1}})$ and ${{x}^{2}}=4a(y-{{m}_{2}})$. We want to find the locus of their point of contact.
Let’s assume $P(x,y)$ is the point of contact of the two given parabolas.
Hence, at this point $P(x,y)$ the slope of tangents for both the parabolas will be the same.
We know that the slope of any curve at a point is equal to the first derivative of the equation of the curve at that point.
So, we will use the formula, slope $=\dfrac{dy}{dx}$ of the equation of the curve.
Now, we will consider the parabola ${{y}^{2}}=4a(x-{{m}_{1}})$.
Differentiating the above equation on both sides, we get $\dfrac{d}{dx}\left( {{y}^{2}} \right)=\dfrac{d}{dx}\left( 4a(x-{{m}_{1}}) \right)$. $...\left( 1 \right)$
To find the value of $\dfrac{d}{dx}\left( {{y}^{2}} \right)$, we can multiply and divide by $dy$. Thus, we have $\dfrac{d}{dx}\left( {{y}^{2}} \right) = \dfrac{d}{dy}\left( {{y}^{2}} \right)\times \dfrac{dy}{dx}$.
We know that differentiation of any function of the form $y=a{{x}^{n}}+b$ is $\dfrac{dy}{dx}=an{{x}^{n-1}}$.
Substituting $a=1,n=2,b=0$ in the above equation, we have $\dfrac{d}{dy}\left( {{y}^{2}} \right)=2y$.
Hence, we get $\dfrac{d}{dx}\left( {{y}^{2}} \right)=\dfrac{d}{dy}\left( {{y}^{2}} \right)\times \dfrac{dy}{dx}=2y\dfrac{dy}{dx}$. $...\left( 2 \right)$
To find the value of $\dfrac{d}{dx}\left( 4a(x-{{m}_{1}}) \right)$, substituting $a=4a,n=1,b=-4a{{m}_{1}}$ in the above equation we get $\dfrac{d}{dx}\left( 4a(x-{{m}_{1}}) \right)=4a$. $...\left( 3 \right)$
Substituting equation $\left( 2 \right)$and$\left( 3 \right)$ in equation $\left( 1 \right)$, we get $\dfrac{d}{dx}\left( {{y}^{2}} \right)=\dfrac{d}{dx}\left( 4a(x-{{m}_{1}}) \right)\Rightarrow 2y\dfrac{dy}{dx}=4a\Rightarrow \dfrac{dy}{dx}=\dfrac{4a}{2y}$.
Thus, we have $\dfrac{dy}{dx}=\dfrac{2a}{y}$. $...\left( 4 \right)$
Now, we will consider the parabola ${{x}^{2}}=4a(y-{{m}_{2}})$.
Differentiating the above equation on both sides, we get $\dfrac{d}{dx}\left( {{x}^{2}} \right)=\dfrac{d}{dx}\left( 4a(y-{{m}_{2}}) \right)$. $...\left( 5 \right)$
To find the value of $\dfrac{d}{dx}\left( 4a\left( y-{{m}_{2}} \right) \right)$, we can multiply and divide by $dy$. Thus, we have $\dfrac{d}{dx}\left( 4a\left( y-{{m}_{2}} \right) \right)=\dfrac{d}{dy}\left( 4a\left( y-{{m}_{2}} \right) \right)\times \dfrac{dy}{dx}$.
We know that differentiation of any function of the form $y=a{{x}^{n}}+b$ is$\dfrac{dy}{dx}=an{{x}^{n-1}}$.
Substituting $a=4a,n=1,b=-4a{{m}_{2}}$ in the above equation, we have $\dfrac{d}{dy}\left( 4a\left( y-{{m}_{2}} \right) \right)=4a$.
Hence, we get $\dfrac{d}{dx}\left( 4a\left( y-{{m}_{2}} \right) \right)=\dfrac{d}{dy}\left( 4a\left( y-{{m}_{2}} \right) \right)\times \dfrac{dy}{dx}=4a\dfrac{dy}{dx}$. $...\left( 6 \right)$
To find the value of $\dfrac{d}{dx}\left( {{x}^{2}} \right)$, substituting $a=1,n=2,b=0$ in the above equation we get $\dfrac{d}{dx}\left( {{x}^{2}} \right)=2x$. $...\left( 7 \right)$
Substituting equation $\left( 6 \right)$and$\left( 7 \right)$ in equation$\left( 5 \right)$, we get$\dfrac{d}{dx}\left( {{x}^{2}} \right)=\dfrac{d}{dx}\left( 4a(y-{{m}_{2}}) \right)\Rightarrow 2x=4a\dfrac{dy}{dx}\Rightarrow \dfrac{dy}{dx}=\dfrac{2x}{4a}$.
Thus, we have $\dfrac{dy}{dx}=\dfrac{x}{2a}$. $...\left( 8 \right)$
As equation $\left( 4 \right)$and$\left( 8 \right)$ represent the same slope, we will equate them to get $\dfrac{dy}{dx}=\dfrac{2a}{y}=\dfrac{x}{2a}$.
Solving this equation, we have $xy=4{{a}^{2}}$.
Hence, the locus of point $P\left( x,y \right)$ which is the point of contact of the two parabolas is a hyperbola.

Note: It’s necessary to use the fact that the slope of tangents at the point of contact of the two parabolas will be the same. We can also write the equation of tangents at the point of contact for the two parabolas and find the slope from those equations of tangents and equate them.