Solveeit Logo
Login

Question

Question: Find the locus of the point P(h, k) if three normals are drawn from the point P to \[{{y}^{2}}=4ax\]...

Find the locus of the point P(h, k) if three normals are drawn from the point P to y2=4ax{{y}^{2}}=4ax, satisfying the following m1.m2=−1{{m}_{1}}.{{m}_{2}}=-1.

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. Doing this we will get our required solution.

Complete step-by-step answer:
We are given a parabola y2=4ax{{y}^{2}}=4ax and a point P(h, k) from which three normals are drawn. Also, given that m1m2=1{{m}_{1}}{{m}_{2}}=1 that is the product of slopes of two out of three normals is -1.
Now, we have to find the locus of the point P(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 the 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 a point (at2,2at)\left( a{{t}^{2}},2at \right) would have 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 = – 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, 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 been given 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 P(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 P(h, k) is (y2−ax+3a2)=0\left( {{y}^{2}}-ax+3{{a}^{2}} \right)=0.

Note: Kindly take care of negative signs 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, take care while taking a2{{a}^{2}} as the LCM.