Question

Find the locus of the point through which three normals to the parabola ${{y}^{2}}=4ax$ pass such that two of them make angles $\alpha$ and $\beta$, respectively, with the axis such that $\tan \alpha \tan \beta =2$.

Answer 1

Hint: First find the equation of the normal which would be a three-degree equation in terms of ‘m’ and put the product of roots = 2.

Complete step-by-step answer:
We are given a parabola ${{y}^{2}}=4ax$ and let P be point P(h, k) from which three normals pass. Also, given that two of them make angles $\alpha$ and $\beta$ with the axis such that $\tan \alpha \tan \beta =2$.
As we know that any line making $\theta$ with the axis would have $m=\tan \theta$ as the slope. Therefore $\tan \alpha$ and $\tan \beta$ are the slopes ${{m}_{1}}$ and ${{m}_{2}}$ of the normals. Therefore, ${{m}_{1}}{{m}_{2}}=2$.
We know that any general point on the parabola ${{y}^{2}}=4ax$ is $\left( x,y \right)=\left( a{{t}^{2}},2at \right)$.
We know that any line passing from $\left( {{x}_{1}},{{y}_{1}} \right)$ and the slope m is:
$\left( y-{{y}_{1}} \right)=m\left( x-{{x}_{1}} \right)$.
So, the equation of normal at the point $\left( a{{t}^{2}},2at \right)$ and the slope m is:
$\left( y-2at \right)=m\left( x-a{{t}^{2}} \right)....\left( i \right)$
Now we take the parabola, ${{y}^{2}}=4ax$.
Now we differentiate the parabola.
$\left[ \text{Also, }\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}} \right]$
Therefore, we get $2y\dfrac{dy}{dx}=4a$
$\dfrac{dy}{dx}=\dfrac{2a}{y}$
At $\left( x,y \right)=\left[ a{{t}^{2}},2at \right]$
We get, $\Rightarrow \dfrac{dy}{dx}=\dfrac{2a}{2at}=\dfrac{1}{t}$
As $\dfrac{dy}{dx}$ signifies the slope of the tangent, therefore any tangent on parabola at a point $\left( a{{t}^{2}},2at \right)$would have a slope $=\dfrac{1}{t}$.
Now, we know that the tangent and normal are perpendicular to each other.
Therefore, $\left( \text{Slope of tangent} \right)\times \left( \text{Slope of normal} \right)=-1$
As we have found that $\text{Slope of tangent =}\dfrac{1}{t}$ and assume that the slope of normal is m.
Therefore, we get $\left( \dfrac{1}{t} \right)\times \left( m \right)=-1$.
Hence, t = – m
Putting the value of t in equation (i), we get,
$\left[ y-2a\left( -m \right) \right]=m\left[ x-a{{\left( -m \right)}^{2}} \right]$
$\Rightarrow y+2am=m\left( x-a{{m}^{2}} \right)$
$\Rightarrow y=mx-a{{m}^{3}}-2am$
By rearranging the given equation, we get,
$a{{m}^{3}}+m\left( 2a-x \right)+y=0$
Here, we get a three-degree equation in terms of m, therefore it will have 3 roots ${{m}_{1}},{{m}_{2}}$ and ${{m}_{3}}$.
As we know that this normal passes through (h, k), we put x = h and y = k. We get,
$a{{m}^{3}}+m\left( 2a-h \right)+k=0....\left( ii \right)$
Comparing the above equation by general three-degree equation $a{{x}^{3}}+b{{x}^{2}}+cx+d=0$
We get, $a=a,\text{ }b=0,\text{ }c=\left( 2a-h \right),\text{ }d=k$
We know that $\text{product of roots }=\dfrac{-d}{a}$
Therefore, we get a product of roots $=\dfrac{-k}{a}$
As ${{m}_{1}},{{m}_{2}}$ and ${{m}_{3}}$ are roots,
Hence, ${{m}_{1}}{{m}_{2}}{{m}_{3}}=\dfrac{-k}{a}$
As we have deduced that ${{m}_{1}}{{m}_{2}}=2$,
We get $\left( 2 \right).{{m}_{3}}=\dfrac{-k}{a}$
Therefore, ${{m}_{3}}=\dfrac{-k}{2a}$
As ${{m}_{3}}$ is the root of the equation (ii), therefore, it will satisfy equation (ii).
Now, we put ${{m}_{3}}$ in equation (ii), we get
$a{{\left( {{m}_{3}} \right)}^{3}}+{{m}_{3}}\left( 2a-h \right)+k=0$
As, ${{m}_{3}}=\dfrac{-k}{2a}$
We get, $a{{\left( \dfrac{-k}{2a} \right)}^{3}}+\left( \dfrac{-k}{2a} \right)\left( 2a-h \right)+k=0$
By simplifying the equation, we get
$\Rightarrow \dfrac{-{{k}^{3}}}{8{{a}^{2}}}+\left( \dfrac{-k}{2a} \right)\left( 2a-h \right)+k=0$
$\Rightarrow \dfrac{-{{k}^{3}}-4ka\left( 2a-h \right)+8k{{a}^{2}}}{8{{a}^{2}}}=0$
$\Rightarrow {{k}^{3}}-4ka\left( 2a-h \right)+8k{{a}^{2}}=0$
By taking k common from each term,
We get, $\left( k \right)\left[ -{{k}^{2}}-4a\left( 2a-h \right)+8{{a}^{2}} \right]=0$
$\Rightarrow \left( k \right)\left[ -{{k}^{2}}-8{{a}^{2}}+4ah+8{{a}^{2}} \right]=0$
Now, to get the locus of P(h, k), we will replace h by x and k by y.
Hence, we get $\Rightarrow \left( y \right)\left( -{{y}^{2}}+4ax \right)=0$
$\Rightarrow -{{y}^{2}}+4ax=0$
So, the locus of point P(h, k) is ${{y}^{2}}=4ax$.

Note: Kindly take care of negative signs with $\dfrac{d}{a}$. Also, many times students confuse with writing the product of roots which is $-\dfrac{d}{a}$ where d is the constant term of the equation. Also, take care while taking ${{a}^{2}}$ as LCM.