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Question: Find the locus of the point N from which 3 normals are drawn to the parabola \[{{y}^{2}}=4ax\] are s...

Find the locus of the point N from which 3 normals are drawn to the parabola y2=4ax{{y}^{2}}=4ax are such that two of them are perpendicular to each other.

Explanation

Solution

Hint: First find the equation of normal which would be a three-degree equation in terms of ‘m’ and put the product of roots -1. By doing this we will get our required solution.

Complete step-by-step answer:
We are given a parabola y2=4ax{{y}^{2}}=4ax and point N(h, k) from which three normals are drawn. Also, given that two of them are perpendicular that means the product of slopes of two out of three normals is – 1.
That means m1m2=−1{{m}_{1}}{{m}_{2}}=-1.
Now, we have to find the locus of the point N(h, k).
We know that any general point on the parabola y2=4ax{{y}^{2}}=4ax is (x,y)=(at2,2at)\left( x,y \right)=\left( a{{t}^{2}},2at \right).
We know that any line passing from (x1,y1)\left( {{x}_{1}},{{y}_{1}} \right) and the slope m is:
(y−y1)=m(x−x1)\left( y-{{y}_{1}} \right)=m\left( x-{{x}_{1}} \right).
So, the equation of normal at the point (at2,2at)\left( a{{t}^{2}},2at \right) and slope m is:
(y−2at)=m(x−at2)....(i)\left( y-2at \right)=m\left( x-a{{t}^{2}} \right)....\left( i \right)
Now we take the parabola, y2=4ax{{y}^{2}}=4ax.
Now we differentiate the parabola.
[Also, d(xn)dx=nxn−1]\left[ \text{Also, }\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}} \right]
Therefore, we get 2ydydx=4a2y\dfrac{dy}{dx}=4a
dydx=2ay\dfrac{dy}{dx}=\dfrac{2a}{y}
At (x,y)=[at2,2at]\left( x,y \right)=\left[ a{{t}^{2}},2at \right]
We get, ⇒dydx=2a2at=1t\Rightarrow \dfrac{dy}{dx}=\dfrac{2a}{2at}=\dfrac{1}{t}
As dydx\dfrac{dy}{dx} signifies the slope of the tangent, therefore any tangent on parabola at the point (at2,2at)\left( a{{t}^{2}},2at \right) would have a slope =1t=\dfrac{1}{t}.
Now, we know that tangent and normal are perpendicular to each other.
Therefore, (Slope of tangent)×(Slope of normal)=−1\left( \text{Slope of tangent} \right)\times \left( \text{Slope of normal} \right)=-1
As we have found that Slope of tangent =1t\text{Slope of tangent =}\dfrac{1}{t} and assume that the slope of normal is m. Therefore, we get,
(1t)×(m)=−1\left( \dfrac{1}{t} \right)\times \left( m \right)=-1.
Hence, t=−mt=-m
Putting the value of t in equation (i), we get,
[y−2a(−m)]=m[x−a(−m)2]\left[ y-2a\left( -m \right) \right]=m\left[ x-a{{\left( -m \right)}^{2}} \right]
⇒y+2am=m(x−am2)\Rightarrow y+2am=m\left( x-a{{m}^{2}} \right)
⇒y=mx−am3−2am\Rightarrow y=mx-a{{m}^{3}}-2am
By rearranging the given equation, we get,
am3+m(2a−x)+y=0a{{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 m1,m2{{m}_{1}},{{m}_{2}} and m3{{m}_{3}}.
As we know that this normal passes through (h, k), we put x = h and y = k. We get,
am3+m(2a−h)+k=0....(ii)a{{m}^{3}}+m\left( 2a-h \right)+k=0....\left( ii \right)
Comparing the above equation by general three-degree equation ax3+bx2+cx+d=0a{{x}^{3}}+b{{x}^{2}}+cx+d=0
We get, a=a, b=0, c=(2a−h), d=ka=a,\text{ }b=0,\text{ }c=\left( 2a-h \right),\text{ }d=k
We know that product of roots =−da\text{product of roots }=\dfrac{-d}{a}
Therefore, we get a product of roots =−ka=\dfrac{-k}{a}
As m1,m2{{m}_{1}},{{m}_{2}} and m3{{m}_{3}} are roots,
Hence, m1m2m3=−ka{{m}_{1}}{{m}_{2}}{{m}_{3}}=\dfrac{-k}{a}
As we have deduced that m1m2=−1{{m}_{1}}{{m}_{2}}=-1,
We get (−1).m3=−ka\left( -1 \right).{{m}_{3}}=\dfrac{-k}{a}
Therefore, m3=ka{{m}_{3}}=\dfrac{k}{a}
As m3{{m}_{3}} is the root of the equation (ii), therefore, it will satisfy equation (ii).
Now, we put m3{{m}_{3}} in equation (ii), we get
a(m3)3+m3(2a−h)+k=0a{{\left( {{m}_{3}} \right)}^{3}}+{{m}_{3}}\left( 2a-h \right)+k=0
As m3=+ka{{m}_{3}}=\dfrac{+k}{a}
We get, a(ka)3+(ka)(2a−h)+k=0a{{\left( \dfrac{k}{a} \right)}^{3}}+\left( \dfrac{k}{a} \right)\left( 2a-h \right)+k=0
By simplifying the equation, we get
⇒k3a2+ka(2a−h)+k=0\Rightarrow \dfrac{{{k}^{3}}}{{{a}^{2}}}+\dfrac{k}{a}\left( 2a-h \right)+k=0
⇒k3+ka(2a−h)+ka2a2=0\Rightarrow \dfrac{{{k}^{3}}+ka\left( 2a-h \right)+k{{a}^{2}}}{{{a}^{2}}}=0
⇒k3+ka(2a−h)+ka2=0\Rightarrow {{k}^{3}}+ka\left( 2a-h \right)+k{{a}^{2}}=0
By taking k common from each term,
We get, (k)[k2+a(2a−h)+a2]=0\left( k \right)\left[ {{k}^{2}}+a\left( 2a-h \right)+{{a}^{2}} \right]=0
⇒(k)[k2+2a2−ah+a2]=0\Rightarrow \left( k \right)\left[ {{k}^{2}}+2{{a}^{2}}-ah+{{a}^{2}} \right]=0
Now, to get the locus of N(h, k), we will replace h by x and k by y.
Hence, we get ⇒(y)(y2−ax+3a2)=0\Rightarrow \left( y \right)\left( {{y}^{2}}-ax+3{{a}^{2}} \right)=0
⇒y2−ax+3a2=0\Rightarrow {{y}^{2}}-ax+3{{a}^{2}}=0
So, the locus of the point N(h, k) is (y2−ax+3a2)=0\left( {{y}^{2}}-ax+3{{a}^{2}} \right)=0.

Note: Kindly take care of the negative sign with da\dfrac{d}{a}. Also, many times students confuse with writing the product of roots which is −da-\dfrac{d}{a} where d is the constant term of the equation. Also, students need to take care while taking a2{{a}^{2}} as LCM.