# A parabola is drawn to pass through \[A\] and \[B\], the ends of a diameter of a given circle of radius \[a\] and to have as directrix a tangent to a concentric circle of radius \[b\]; the axes being \[AB\] and a perpendicular diameter, prove that the locus of the focus of the parabola is \[\dfrac{{{x}^{2}}}{{{b}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}-{{a}^{2}}}=1\].

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Hint: Any point on the parabola is equidistant from the focus and directrix.

Let the equation of the circle be \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}.....\left( i \right)\]

In the question, it is given that \[A\] and \[B\] are the ends of the diameter of the circle.

So, let \[A=\left( a,0 \right)\] and \[B=\left( -a,0 \right)\]

In the question, it is given that the radius of the concentric circle is \[b\].

So, let the equation of concentric circle be \[{{x}^{2}}+{{y}^{2}}={{b}^{2}}....\left( ii \right)\]

Now, the directrix of the parabola is tangent to the concentric circle.

First, we find the equation of tangent to the concentric circle.

We know, the equation of tangent to a circle \[{{x}^{2}}+{{y}^{2}}={{r}^{2}}\] in slope form is given as \[y=mx\pm r\sqrt{1+{{m}^{2}}}\].

So, equation of tangent to \[\left( ii \right)\]is

\[y=mx+b\sqrt{{{m}^{2}}+1}\]

Or \[mx-y+b\sqrt{{{m}^{2}}+1}=0....\left( iii \right)\]

Equation \[\left( iii \right)\] is also the directrix to the parabola.

Now, we want to find the locus of the focus of the parabola.

So, let the focus of the parabola be \[F\left( h,k \right)\].

Now, by the definition of a parabola, any point on the parabola is equidistant from the focus and the directrix.

Now, from the question, we know that the ends of diameter of \[\left( i \right)\]i.e. \[A\] and \[B\] lie on the parabola.

We know that the distance between two points \[({{x}_{1}},{{y}_{1}})\] and \[({{x}_{2}},{{y}_{2}})\] is given as \[d=\sqrt{{{({{x}_{1}}-{{x}_{2}})}^{2}}+{{({{y}_{1}}-{{y}_{2}})}^{2}}}\] and the distance of a point \[({{x}_{1}},{{y}_{1}})\] from the line \[lx+my+n=0\] is given

as \[d=\dfrac{l{{x}_{1}}+m{{y}_{1}}+n}{\sqrt{{{l}^{2}}+{{m}^{2}}}}\].

So, for \[A\left( a,0 \right)\], we have

\[\sqrt{{{\left( h-a \right)}^{2}}+{{k}^{2}}}=\dfrac{m\left( a \right)-1\left( 0 \right)+b\sqrt{{{m}^{2}}+1}}{\sqrt{{{m}^{2}}+1}}\]

\[\Rightarrow \sqrt{{{\left( h-a \right)}^{2}}+{{k}^{2}}}=\dfrac{am+b\sqrt{{{m}^{2}}+1}}{\sqrt{{{m}^{2}}+1}}....\left( iv \right)\]

For \[B\left( -a,0 \right)\], we have

\[\sqrt{{{\left( h+a \right)}^{2}}+{{k}^{2}}}=\dfrac{m\left( -a \right)-1\left( 0 \right)+b\sqrt{{{m}^{2}}+1}}{\sqrt{{{m}^{2}}+1}}\]

\[\Rightarrow \sqrt{{{\left( h+a \right)}^{2}}+{{k}^{2}}}=\dfrac{-am+b\sqrt{{{m}^{2}}+1}}{\sqrt{{{m}^{2}}+1}}....\left( v \right)\]

Adding \[\left( iv \right)\]and \[\left( v \right)\], we have

\[\sqrt{{{\left( h-a \right)}^{2}}+{{k}^{2}}}+\sqrt{{{\left( h+a \right)}^{2}}+{{k}^{2}}}=\dfrac{2b\sqrt{{{m}^{2}}+1}}{\sqrt{{{m}^{2}}+1}}\]

\[\Rightarrow \sqrt{{{\left( h-a \right)}^{2}}+{{k}^{2}}}+\sqrt{{{\left( h+a \right)}^{2}}+{{k}^{2}}}=2b\]

\[\sqrt{{{\left( h+a \right)}^{2}}+{{k}^{2}}}=2b-\sqrt{{{\left( h-a \right)}^{2}}+{{k}^{2}}}\]

Now, on squaring both sides we get,

\[{{h}^{2}}+2ah+{{a}^{2}}+{{k}^{2}}=4{{b}^{2}}+{{h}^{2}}-2ah+{{a}^{2}}+{{k}^{2}}-4b\sqrt{{{(h-a)}^{2}}+{{k}^{2}}}\]

\[\begin{align}

& \Rightarrow 4{{b}^{2}}-4ah=4b\sqrt{{{(h-a)}^{2}}+{{k}^{2}}} \\

& \Rightarrow {{b}^{2}}-ah=b\sqrt{{{(h-a)}^{2}}+{{k}^{2}}} \\

\end{align}\]

Now, let’s square both sides again to remove the square root sign. We get \[{{b}^{4}}+{{a}^{2}}{{h}^{2}}-2{{b}^{2}}ah={{b}^{2}}({{h}^{2}}-2ah+{{a}^{2}}+{{k}^{2}})\]

On dividing both sides with \[{{b}^{2}}\], we get

\[{{b}^{2}}+\dfrac{{{a}^{2}}{{h}^{2}}}{{{b}^{2}}}-2ah={{h}^{2}}-2ah+{{a}^{2}}+{{k}^{2}}\]

\[\Rightarrow {{b}^{2}}-{{a}^{2}}-{{k}^{2}}={{h}^{2}}-\dfrac{{{a}^{2}}{{h}^{2}}}{{{b}^{2}}}\]

\[\Rightarrow ({{b}^{2}}-{{a}^{2}})(1-\dfrac{{{k}^{2}}}{{{b}^{2}}-{{a}^{2}}})={{h}^{2}}(1-\dfrac{{{a}^{2}}}{{{b}^{2}}})\]

\[\Rightarrow ({{b}^{2}}-{{a}^{2}})(1-\dfrac{{{k}^{2}}}{{{b}^{2}}-{{a}^{2}}})={{h}^{2}}(\dfrac{{{b}^{2}}-{{a}^{2}}}{{{b}^{2}}})\]

\[\Rightarrow 1-\dfrac{{{k}^{2}}}{{{b}^{2}}-{{a}^{2}}}=\dfrac{{{h}^{2}}}{{{b}^{2}}}\]

\[\Rightarrow 1=\dfrac{{{h}^{2}}}{{{b}^{2}}}+\dfrac{{{k}^{2}}}{{{b}^{2}}-{{a}^{2}}}\]

So, the locus of \[(h,k)\] is given by replacing \[(h,k)\]by \[(x,y)\]

So, the equation of the locus is given as \[\dfrac{{{x}^{2}}}{{{b}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}-{{a}^{2}}}=1\] \[\]\[\]

Note: The distance between two points \[({{x}_{1}},{{y}_{1}})\] and \[({{x}_{2}},{{y}_{2}})\] is given as \[d=\sqrt{{{({{x}_{1}}-{{x}_{2}})}^{2}}+{{({{y}_{1}}-{{y}_{2}})}^{2}}}\] and not \[d=\sqrt{{{({{x}_{1}}+{{x}_{2}})}^{2}}+{{({{y}_{1}}+{{y}_{2}})}^{2}}}\]. It is a very common mistake made by students.

Let the equation of the circle be \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}.....\left( i \right)\]

In the question, it is given that \[A\] and \[B\] are the ends of the diameter of the circle.

So, let \[A=\left( a,0 \right)\] and \[B=\left( -a,0 \right)\]

In the question, it is given that the radius of the concentric circle is \[b\].

So, let the equation of concentric circle be \[{{x}^{2}}+{{y}^{2}}={{b}^{2}}....\left( ii \right)\]

Now, the directrix of the parabola is tangent to the concentric circle.

First, we find the equation of tangent to the concentric circle.

We know, the equation of tangent to a circle \[{{x}^{2}}+{{y}^{2}}={{r}^{2}}\] in slope form is given as \[y=mx\pm r\sqrt{1+{{m}^{2}}}\].

So, equation of tangent to \[\left( ii \right)\]is

\[y=mx+b\sqrt{{{m}^{2}}+1}\]

Or \[mx-y+b\sqrt{{{m}^{2}}+1}=0....\left( iii \right)\]

Equation \[\left( iii \right)\] is also the directrix to the parabola.

Now, we want to find the locus of the focus of the parabola.

So, let the focus of the parabola be \[F\left( h,k \right)\].

Now, by the definition of a parabola, any point on the parabola is equidistant from the focus and the directrix.

Now, from the question, we know that the ends of diameter of \[\left( i \right)\]i.e. \[A\] and \[B\] lie on the parabola.

We know that the distance between two points \[({{x}_{1}},{{y}_{1}})\] and \[({{x}_{2}},{{y}_{2}})\] is given as \[d=\sqrt{{{({{x}_{1}}-{{x}_{2}})}^{2}}+{{({{y}_{1}}-{{y}_{2}})}^{2}}}\] and the distance of a point \[({{x}_{1}},{{y}_{1}})\] from the line \[lx+my+n=0\] is given

as \[d=\dfrac{l{{x}_{1}}+m{{y}_{1}}+n}{\sqrt{{{l}^{2}}+{{m}^{2}}}}\].

So, for \[A\left( a,0 \right)\], we have

\[\sqrt{{{\left( h-a \right)}^{2}}+{{k}^{2}}}=\dfrac{m\left( a \right)-1\left( 0 \right)+b\sqrt{{{m}^{2}}+1}}{\sqrt{{{m}^{2}}+1}}\]

\[\Rightarrow \sqrt{{{\left( h-a \right)}^{2}}+{{k}^{2}}}=\dfrac{am+b\sqrt{{{m}^{2}}+1}}{\sqrt{{{m}^{2}}+1}}....\left( iv \right)\]

For \[B\left( -a,0 \right)\], we have

\[\sqrt{{{\left( h+a \right)}^{2}}+{{k}^{2}}}=\dfrac{m\left( -a \right)-1\left( 0 \right)+b\sqrt{{{m}^{2}}+1}}{\sqrt{{{m}^{2}}+1}}\]

\[\Rightarrow \sqrt{{{\left( h+a \right)}^{2}}+{{k}^{2}}}=\dfrac{-am+b\sqrt{{{m}^{2}}+1}}{\sqrt{{{m}^{2}}+1}}....\left( v \right)\]

Adding \[\left( iv \right)\]and \[\left( v \right)\], we have

\[\sqrt{{{\left( h-a \right)}^{2}}+{{k}^{2}}}+\sqrt{{{\left( h+a \right)}^{2}}+{{k}^{2}}}=\dfrac{2b\sqrt{{{m}^{2}}+1}}{\sqrt{{{m}^{2}}+1}}\]

\[\Rightarrow \sqrt{{{\left( h-a \right)}^{2}}+{{k}^{2}}}+\sqrt{{{\left( h+a \right)}^{2}}+{{k}^{2}}}=2b\]

\[\sqrt{{{\left( h+a \right)}^{2}}+{{k}^{2}}}=2b-\sqrt{{{\left( h-a \right)}^{2}}+{{k}^{2}}}\]

Now, on squaring both sides we get,

\[{{h}^{2}}+2ah+{{a}^{2}}+{{k}^{2}}=4{{b}^{2}}+{{h}^{2}}-2ah+{{a}^{2}}+{{k}^{2}}-4b\sqrt{{{(h-a)}^{2}}+{{k}^{2}}}\]

\[\begin{align}

& \Rightarrow 4{{b}^{2}}-4ah=4b\sqrt{{{(h-a)}^{2}}+{{k}^{2}}} \\

& \Rightarrow {{b}^{2}}-ah=b\sqrt{{{(h-a)}^{2}}+{{k}^{2}}} \\

\end{align}\]

Now, let’s square both sides again to remove the square root sign. We get \[{{b}^{4}}+{{a}^{2}}{{h}^{2}}-2{{b}^{2}}ah={{b}^{2}}({{h}^{2}}-2ah+{{a}^{2}}+{{k}^{2}})\]

On dividing both sides with \[{{b}^{2}}\], we get

\[{{b}^{2}}+\dfrac{{{a}^{2}}{{h}^{2}}}{{{b}^{2}}}-2ah={{h}^{2}}-2ah+{{a}^{2}}+{{k}^{2}}\]

\[\Rightarrow {{b}^{2}}-{{a}^{2}}-{{k}^{2}}={{h}^{2}}-\dfrac{{{a}^{2}}{{h}^{2}}}{{{b}^{2}}}\]

\[\Rightarrow ({{b}^{2}}-{{a}^{2}})(1-\dfrac{{{k}^{2}}}{{{b}^{2}}-{{a}^{2}}})={{h}^{2}}(1-\dfrac{{{a}^{2}}}{{{b}^{2}}})\]

\[\Rightarrow ({{b}^{2}}-{{a}^{2}})(1-\dfrac{{{k}^{2}}}{{{b}^{2}}-{{a}^{2}}})={{h}^{2}}(\dfrac{{{b}^{2}}-{{a}^{2}}}{{{b}^{2}}})\]

\[\Rightarrow 1-\dfrac{{{k}^{2}}}{{{b}^{2}}-{{a}^{2}}}=\dfrac{{{h}^{2}}}{{{b}^{2}}}\]

\[\Rightarrow 1=\dfrac{{{h}^{2}}}{{{b}^{2}}}+\dfrac{{{k}^{2}}}{{{b}^{2}}-{{a}^{2}}}\]

So, the locus of \[(h,k)\] is given by replacing \[(h,k)\]by \[(x,y)\]

So, the equation of the locus is given as \[\dfrac{{{x}^{2}}}{{{b}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}-{{a}^{2}}}=1\] \[\]\[\]

Note: The distance between two points \[({{x}_{1}},{{y}_{1}})\] and \[({{x}_{2}},{{y}_{2}})\] is given as \[d=\sqrt{{{({{x}_{1}}-{{x}_{2}})}^{2}}+{{({{y}_{1}}-{{y}_{2}})}^{2}}}\] and not \[d=\sqrt{{{({{x}_{1}}+{{x}_{2}})}^{2}}+{{({{y}_{1}}+{{y}_{2}})}^{2}}}\]. It is a very common mistake made by students.

Last updated date: 03rd Jun 2023

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