The coordinates of the point on the parabola 24xy which is nearest to the circle
\[{{(x-3)}^{2}}+{{y}^{2}}=1\] are:
(a) (0,0)
(b) (2,1)
(c) (2,1)
(d) (4,4)
Answer
328.8k+ views
Hint: To find the coordinates of the point on the parabola which has the least
distance from the circle, we will find the distance between the centre of the circle and any
point on the parabola taken in parametric form. Then, we will differentiate that distance and
equate it to zero to get the coordinates of the point which is at minimum distance from the
circle.
We have a parabola\[{{x}^{2}}=4y\] and a circle\[{{(x-3)}^{2}}+{{y}^{2}}=1\]. We want to find
the coordinates of the point on the parabola which has the least distance from the circle.
We know that any general point on the parabola of the
form\[{{x}^{2}}=4ay\]is\[(2at,a{{t}^{2}})\].
By substituting\[a=1\], we have the point on our parabola as\[(2t,{{t}^{2}})\].
We know that a circle of the form\[{{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}\]has centre\[(h,k)\].
Thus the centre of the circle\[{{(x-3)}^{2}}+{{y}^{2}}=1\]is\[(3,0)\].
We know that the distance between any two points \[({{x}_{1}},{{y}_{1}})\] and \[({{x}_{2}},{{y}_{2}})\] is \[\sqrt{{({x_1}-{x_2})}^2+{({y_1}-{y_2})}^2}\].
Substituting \[{{x}_{1}}=2t,{{y}_{1}}={{t}^{2}},{{x}_{2}}=3,{{y}_{2}}=0\] in the above formula,
we get the distance between the points to be\[\sqrt{{{(2t-3)}^{2}}+{{t}^{4}}}\].
We know that to find the minimum value of a function\[f(x)\], we have\[\dfrac{df(x)}{dx}=0\].
So, we will differentiate our function \[\sqrt{{{(2t-3)}^{2}}+{{t}^{4}}}\] with respect to\[t\].
Thus, we get
\[\Rightarrow \dfrac{d}{dt}\sqrt{{{(2t-3)}^{2}}+{{t}^{4}}}=\dfrac{1}{2\sqrt{{{(2t-3)}^{2}}+{{t}^{4}}}}\{2(2t-3)2+
4{{t}^{3}}\}=0\]
\[\begin{align}
& \Rightarrow 4(2t-3)+4{{t}^{3}}=0 \\
& \Rightarrow 8t-12+4{{t}^{3}}=0 \\
\end{align}\]
We will now factorize the above equation. Add and subtract\[4{{t}^{2}}\]from the equation.
Thus we have\[4{{t}^{3}}+8t-12=4{{t}^{3}}-4{{t}^{2}}+4{{t}^{2}}+8t-12\].
\[\begin{align}
& \Rightarrow 4{{t}^{3}}-4{{t}^{2}}+4{{t}^{2}}+8t-12=0 \\
& \Rightarrow 4{{t}^{2}}(t-1)+4({{t}^{2}}+2t-3)=0 \\
& \Rightarrow 4{{t}^{2}}(t-1)+4({{t}^{2}}+3t-t-3)=0 \\
& \Rightarrow 4{{t}^{2}}(t-1)+4(t(t+3)-1(t+3))=0 \\
& \Rightarrow 4{{t}^{2}}(t-1)+4(t-1)(t+3)=0 \\
& \Rightarrow 4(t-1)({{t}^{2}}+t+3)=0 \\
\end{align}\]
We observe that\[t=1\]satisfies the equation and the equation\[{{t}^{2}}+t+3\]has no real
roots.
Substituting the value of\[t\]in the point\[(2t,{{t}^{2}})\], we get\[(2,1)\].
So, the point on the parabola which is at minimum distance from the circle is (2,1).
Hence, the correct answer is \[(2,1)\].
Note: We can also solve this question by finding the foot of perpendicular from any general
point on parabola to the circle and then differentiating the distance between the foot of
perpendicular on circle and point on parabola and equating it to zero.
distance from the circle, we will find the distance between the centre of the circle and any
point on the parabola taken in parametric form. Then, we will differentiate that distance and
equate it to zero to get the coordinates of the point which is at minimum distance from the
circle.
We have a parabola\[{{x}^{2}}=4y\] and a circle\[{{(x-3)}^{2}}+{{y}^{2}}=1\]. We want to find
the coordinates of the point on the parabola which has the least distance from the circle.
We know that any general point on the parabola of the
form\[{{x}^{2}}=4ay\]is\[(2at,a{{t}^{2}})\].
By substituting\[a=1\], we have the point on our parabola as\[(2t,{{t}^{2}})\].

We know that a circle of the form\[{{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}\]has centre\[(h,k)\].
Thus the centre of the circle\[{{(x-3)}^{2}}+{{y}^{2}}=1\]is\[(3,0)\].
We know that the distance between any two points \[({{x}_{1}},{{y}_{1}})\] and \[({{x}_{2}},{{y}_{2}})\] is \[\sqrt{{({x_1}-{x_2})}^2+{({y_1}-{y_2})}^2}\].
Substituting \[{{x}_{1}}=2t,{{y}_{1}}={{t}^{2}},{{x}_{2}}=3,{{y}_{2}}=0\] in the above formula,
we get the distance between the points to be\[\sqrt{{{(2t-3)}^{2}}+{{t}^{4}}}\].
We know that to find the minimum value of a function\[f(x)\], we have\[\dfrac{df(x)}{dx}=0\].
So, we will differentiate our function \[\sqrt{{{(2t-3)}^{2}}+{{t}^{4}}}\] with respect to\[t\].
Thus, we get
\[\Rightarrow \dfrac{d}{dt}\sqrt{{{(2t-3)}^{2}}+{{t}^{4}}}=\dfrac{1}{2\sqrt{{{(2t-3)}^{2}}+{{t}^{4}}}}\{2(2t-3)2+
4{{t}^{3}}\}=0\]
\[\begin{align}
& \Rightarrow 4(2t-3)+4{{t}^{3}}=0 \\
& \Rightarrow 8t-12+4{{t}^{3}}=0 \\
\end{align}\]
We will now factorize the above equation. Add and subtract\[4{{t}^{2}}\]from the equation.
Thus we have\[4{{t}^{3}}+8t-12=4{{t}^{3}}-4{{t}^{2}}+4{{t}^{2}}+8t-12\].
\[\begin{align}
& \Rightarrow 4{{t}^{3}}-4{{t}^{2}}+4{{t}^{2}}+8t-12=0 \\
& \Rightarrow 4{{t}^{2}}(t-1)+4({{t}^{2}}+2t-3)=0 \\
& \Rightarrow 4{{t}^{2}}(t-1)+4({{t}^{2}}+3t-t-3)=0 \\
& \Rightarrow 4{{t}^{2}}(t-1)+4(t(t+3)-1(t+3))=0 \\
& \Rightarrow 4{{t}^{2}}(t-1)+4(t-1)(t+3)=0 \\
& \Rightarrow 4(t-1)({{t}^{2}}+t+3)=0 \\
\end{align}\]
We observe that\[t=1\]satisfies the equation and the equation\[{{t}^{2}}+t+3\]has no real
roots.
Substituting the value of\[t\]in the point\[(2t,{{t}^{2}})\], we get\[(2,1)\].
So, the point on the parabola which is at minimum distance from the circle is (2,1).
Hence, the correct answer is \[(2,1)\].
Note: We can also solve this question by finding the foot of perpendicular from any general
point on parabola to the circle and then differentiating the distance between the foot of
perpendicular on circle and point on parabola and equating it to zero.
Last updated date: 30th May 2023
•
Total views: 328.8k
•
Views today: 2.86k
Recently Updated Pages
Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE
