# Find the range of \['a'\] for which the parabola \[y=a{{x}^{2}}\] and the unit circle with centre at \[(0,1)\] meet each other at two points other than origin.

Last updated date: 19th Mar 2023

•

Total views: 305.4k

•

Views today: 2.84k

Answer

Verified

305.4k+ views

Hint: To find the range of \['a'\] for which the parabola and the circle intersect each other at points other than origin, we will first find the equation of the circle with the given centre and then substitute the equation of the parabola in the equation of the circle and solve it to get the desired range.

Complete step-by-step answer:

We have a parabola of the form \[y=a{{x}^{2}}\] and a unit circle with centre at \[(0,1)\].

We want to find the range of \['a'\] for which the two curves will intersect at points other than the origin.

We know that the equation of circle with centre \[(h,k)\] and radius \[r\] is\[{{(x-h)}^{2}}+{{(y-

k)}^{2}}={{r}^{2}}\].

Substituting \[h=0,k=1,r=1\] in the above equation, we get \[{{x}^{2}}+{{(y-1)}^{2}}={{1}^{2}}\]

\[...(1)\] as the equation of our circle.

Now, we have \[y=a{{x}^{2}}\] \[...(2)\] as the equation of our parabola.

To find the point of intersection of the two curves, we will rewrite the equation of parabola

as \[\dfrac{y}{a}={{x}^{2}}\] \[...(3)\]

We will substitute this equation of parabola in equation \[(1)\].

Thus, we get

\[\begin{align}

&\Rightarrow\dfrac{y}{a}+{{(y-1)^{2}}}={{1}^{2}}\\

&\Rightarrow\dfrac{y}{a}+{{y}^{2}}+1-2y=1

\\ &\Rightarrow\dfrac{y}{a}+{{y}^{2}}-2y=0

\end{align}\]

Now factorizing the above equation, we get \[y(\dfrac{1}{a}+y-2)=0\].

So, we have

\[\Rightarrow y=0,y=2-\dfrac{1}{a}\]

Thus, for a point of intersection of two curves other than the origin, we have \[y=2-\dfrac{1}{a}\].

Substituting the above value of \[y\] in the equation \[(3)\], we get \[{{x}^{2}}=\dfrac{y}{a}=\dfrac{2}{a}-\dfrac{1}{{{a}^{2}}}\].

As we are excluding the origin from our point of intersection of the two curves, we have \[{{x}^{2}}>0\]

\[\begin{align}

& \Rightarrow \dfrac{2}{a}-\dfrac{1}{{{a}^{2}}}>0 \\

& \Rightarrow \dfrac{2a-1}{{{a}^{2}}}>0 \\

& \Rightarrow 2a-1>0 \\

& \Rightarrow a>\dfrac{1}{2} \\

\end{align}\]

\[\Rightarrow a\in \left( \dfrac{1}{2},\infty \right)\]

Hence, the required range of value of \['a'\] is \[a\in \left( \dfrac{1}{2},\infty \right)\].

Note: We can also solve this question by substituting the value of \[y\] in terms of \[x\] and then solve for \[x\]. However, that will be time consuming to calculate. Also, one must keep in mind that \[{{x}^{2}}>0\] as we are looking for points of intersection of two curves other than the origin.

Complete step-by-step answer:

We have a parabola of the form \[y=a{{x}^{2}}\] and a unit circle with centre at \[(0,1)\].

We want to find the range of \['a'\] for which the two curves will intersect at points other than the origin.

We know that the equation of circle with centre \[(h,k)\] and radius \[r\] is\[{{(x-h)}^{2}}+{{(y-

k)}^{2}}={{r}^{2}}\].

Substituting \[h=0,k=1,r=1\] in the above equation, we get \[{{x}^{2}}+{{(y-1)}^{2}}={{1}^{2}}\]

\[...(1)\] as the equation of our circle.

Now, we have \[y=a{{x}^{2}}\] \[...(2)\] as the equation of our parabola.

To find the point of intersection of the two curves, we will rewrite the equation of parabola

as \[\dfrac{y}{a}={{x}^{2}}\] \[...(3)\]

We will substitute this equation of parabola in equation \[(1)\].

Thus, we get

\[\begin{align}

&\Rightarrow\dfrac{y}{a}+{{(y-1)^{2}}}={{1}^{2}}\\

&\Rightarrow\dfrac{y}{a}+{{y}^{2}}+1-2y=1

\\ &\Rightarrow\dfrac{y}{a}+{{y}^{2}}-2y=0

\end{align}\]

Now factorizing the above equation, we get \[y(\dfrac{1}{a}+y-2)=0\].

So, we have

\[\Rightarrow y=0,y=2-\dfrac{1}{a}\]

Thus, for a point of intersection of two curves other than the origin, we have \[y=2-\dfrac{1}{a}\].

Substituting the above value of \[y\] in the equation \[(3)\], we get \[{{x}^{2}}=\dfrac{y}{a}=\dfrac{2}{a}-\dfrac{1}{{{a}^{2}}}\].

As we are excluding the origin from our point of intersection of the two curves, we have \[{{x}^{2}}>0\]

\[\begin{align}

& \Rightarrow \dfrac{2}{a}-\dfrac{1}{{{a}^{2}}}>0 \\

& \Rightarrow \dfrac{2a-1}{{{a}^{2}}}>0 \\

& \Rightarrow 2a-1>0 \\

& \Rightarrow a>\dfrac{1}{2} \\

\end{align}\]

\[\Rightarrow a\in \left( \dfrac{1}{2},\infty \right)\]

Hence, the required range of value of \['a'\] is \[a\in \left( \dfrac{1}{2},\infty \right)\].

Note: We can also solve this question by substituting the value of \[y\] in terms of \[x\] and then solve for \[x\]. However, that will be time consuming to calculate. Also, one must keep in mind that \[{{x}^{2}}>0\] as we are looking for points of intersection of two curves other than the origin.

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