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(A) -1 < m < 1

(B) m< -1 or m > 1

(C) $m\in R$

(D) None of the above

Answer
Verified

Hint: Focus of parabola with equation ${{y}^{2}}=4\left( x-1 \right)$is (a, 0). Quadratic $A{{x}^{2}}+Bx+C=0$have real roots if ${{B}^{2}}=4AC=$Discriminant is greater than zero.

Complete step-by-step answer:

The given equation of parabola is

${{y}^{2}}=4\left( x-1 \right)............\left( 1 \right)$

As we know, the standard equation of parabola is ${{y}^{2}}=4ax$ with vertex (0, 0) and focus (a, 0). We can write the standard equation as ${{\left( y-0 \right)}^{2}}=4a\left( x-0 \right)$. Now, generalizing the above relation, we can rewrite parabola equation as ${{\left( y-{{y}_{1}} \right)}^{2}}=4a\left( x-{{x}_{1}} \right)$ where $\left( {{x}_{1}}-{{y}_{1}} \right)$are vertex of the parabola and focus will shift to $\left( a+{{x}_{1}},{{y}_{1}} \right)$.

Now, comparing equation ${{\left( y-{{y}_{1}} \right)}^{2}}=4a\left( x-{{x}_{1}} \right)$ with the given parabola equation (1) i.e. \[{{y}^{2}}=4\left( x-1 \right)or{{\left( y-0 \right)}^{2}}=4\left( x-1 \right)\], we get;

Vertex $=\left( {{x}_{1}},{{y}_{1}} \right)=\left( 1,0 \right)$

Focus $=\left( 1+1,0 \right)=\left( 2,0 \right)$

We can show vertex and focus of given parabola in diagram given below;

Now, we have line L passing through focus and intersecting the given parabola at two points.

Let us suppose line L as

$y=mx+c......\left( 2 \right)$

Where m is the slope of the line and c is the intercept on y-axis.

As, equation (2) is passing through focus (2, 0). Hence, it will satisfy the equation (2),

$\begin{align}

& 0=2m+c \\

& c=-2m..........\left( 3 \right) \\

\end{align}$

Now, equation (2) can be rewritten in variable â€˜mâ€™ as

$y=mx-2m...............\left( 4 \right)$

As we are given that equation (4) is intersecting at two points with the given parabola as expressed equation (1).

Now, for getting intersecting points, we can substitute â€˜yâ€™ from equation (4) to equation (1) i.e.${{y}^{2}}=4\left( x-1 \right)$. Hence, we get

${{\left( mx-2m \right)}^{2}}=4\left( x-1 \right)$

Simplifying \[{{\left( mx-2m \right)}^{2}}\] by using algebraic identity \[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\], we get

\[\begin{align}

& {{m}^{2}}{{x}^{2}}+4{{m}^{2}}-4{{m}^{2}}x=4x-4 \\

& Or \\

& {{m}^{2}}{{x}^{2}}-x(4{{m}^{2}}+4)+4{{m}^{2}}+4=0..........\left( 5 \right) \\

\end{align}\]

As the above equation is a quadratic equation in â€˜xâ€™, so we will get two roots of equation (5) which are x-coordinates of both the intersecting points of line â€˜Lâ€™ with the given parabola.

As both the roots should be real.

Now, we know that any quadratic $A{{x}^{2}}+Bx+C=0$will have both roots real, if discriminant or $D={{b}^{2}}-4ac$ is greater than zero.

Hence, equation (5) has both roots real so discriminant of it should be greater than 0. Hence, we get,

$\begin{align}

& {{\left( 4{{m}^{2}}+4 \right)}^{2}}-4{{m}^{2}}\left( 4{{m}^{2}}+4 \right)>0 \\

& Or \\

& 16{{m}^{4}}+16+64{{m}^{2}}-16{{m}^{4}}-16{{m}^{2}}>0 \\

\end{align}$

On simplifying the above relation, we get,

$48{{m}^{2}}+16>0..........\left( 6 \right)$

Now, we can observe that $48{{m}^{2}}+16$ will always be greater than zero for any real value of â€˜mâ€™. ${{m}^{2}}$will never be negative, for any real value of â€˜mâ€™. Hence, $m\in R$ is the condition for the given question that line â€˜Lâ€™ always passes through two points if it passes through focus of given parabola.

Hence, option (C) is the correct answer.

Note: One can go wrong while writing the focus of the parabola of a given equation. He/she may take (1, 0) as focus of the given parabola. But it will be wrong and need to compare with ${{\left( y-{{y}_{1}} \right)}^{2}}=4a\left( x-{{x}_{1}} \right)$ where vertex is $\left( {{x}_{1}},{{y}_{1}} \right)$and focus is $\left( {{x}_{1}}+a,{{y}_{1}} \right)$.

Hence, be careful while writing focus or vertex in these kinds of questions.

Another approach for this question would be that we can suppose two parametric coordinates on the given parabola as, \[\left( t_{1}^{2}+1,2{{t}_{1}} \right)\ and\ \left( t_{2}^{2}+1,2{{t}_{2}} \right)\]. A line is passing through (2,0) as well.

Hence, three points are collinear. Now, we know that if $\left( {{x}_{1}},{{y}_{1}} \right)$, $\left( {{x}_{2}},{{y}_{2}} \right)\ and\ \left( {{x}_{3}},{{y}_{3}} \right)$ are collinear then triangle formed by all three points have area 0. Hence,

$\left| \begin{matrix}

{{x}_{1}} & {{y}_{1}} & 1 \\

{{x}_{2}} & {{y}_{2}} & 1 \\

{{x}_{3}} & {{y}_{3}} & 1 \\

\end{matrix} \right|=0$

And slope can be given as $\dfrac{2\left( {{t}_{2}}-{{t}_{1}} \right)}{t_{2}^{2}-t_{1}^{2}}$ which is equal to m. Now, using both the equations, try to get a relation between â€˜mâ€™.

Complete step-by-step answer:

The given equation of parabola is

${{y}^{2}}=4\left( x-1 \right)............\left( 1 \right)$

As we know, the standard equation of parabola is ${{y}^{2}}=4ax$ with vertex (0, 0) and focus (a, 0). We can write the standard equation as ${{\left( y-0 \right)}^{2}}=4a\left( x-0 \right)$. Now, generalizing the above relation, we can rewrite parabola equation as ${{\left( y-{{y}_{1}} \right)}^{2}}=4a\left( x-{{x}_{1}} \right)$ where $\left( {{x}_{1}}-{{y}_{1}} \right)$are vertex of the parabola and focus will shift to $\left( a+{{x}_{1}},{{y}_{1}} \right)$.

Now, comparing equation ${{\left( y-{{y}_{1}} \right)}^{2}}=4a\left( x-{{x}_{1}} \right)$ with the given parabola equation (1) i.e. \[{{y}^{2}}=4\left( x-1 \right)or{{\left( y-0 \right)}^{2}}=4\left( x-1 \right)\], we get;

Vertex $=\left( {{x}_{1}},{{y}_{1}} \right)=\left( 1,0 \right)$

Focus $=\left( 1+1,0 \right)=\left( 2,0 \right)$

We can show vertex and focus of given parabola in diagram given below;

Now, we have line L passing through focus and intersecting the given parabola at two points.

Let us suppose line L as

$y=mx+c......\left( 2 \right)$

Where m is the slope of the line and c is the intercept on y-axis.

As, equation (2) is passing through focus (2, 0). Hence, it will satisfy the equation (2),

$\begin{align}

& 0=2m+c \\

& c=-2m..........\left( 3 \right) \\

\end{align}$

Now, equation (2) can be rewritten in variable â€˜mâ€™ as

$y=mx-2m...............\left( 4 \right)$

As we are given that equation (4) is intersecting at two points with the given parabola as expressed equation (1).

Now, for getting intersecting points, we can substitute â€˜yâ€™ from equation (4) to equation (1) i.e.${{y}^{2}}=4\left( x-1 \right)$. Hence, we get

${{\left( mx-2m \right)}^{2}}=4\left( x-1 \right)$

Simplifying \[{{\left( mx-2m \right)}^{2}}\] by using algebraic identity \[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\], we get

\[\begin{align}

& {{m}^{2}}{{x}^{2}}+4{{m}^{2}}-4{{m}^{2}}x=4x-4 \\

& Or \\

& {{m}^{2}}{{x}^{2}}-x(4{{m}^{2}}+4)+4{{m}^{2}}+4=0..........\left( 5 \right) \\

\end{align}\]

As the above equation is a quadratic equation in â€˜xâ€™, so we will get two roots of equation (5) which are x-coordinates of both the intersecting points of line â€˜Lâ€™ with the given parabola.

As both the roots should be real.

Now, we know that any quadratic $A{{x}^{2}}+Bx+C=0$will have both roots real, if discriminant or $D={{b}^{2}}-4ac$ is greater than zero.

Hence, equation (5) has both roots real so discriminant of it should be greater than 0. Hence, we get,

$\begin{align}

& {{\left( 4{{m}^{2}}+4 \right)}^{2}}-4{{m}^{2}}\left( 4{{m}^{2}}+4 \right)>0 \\

& Or \\

& 16{{m}^{4}}+16+64{{m}^{2}}-16{{m}^{4}}-16{{m}^{2}}>0 \\

\end{align}$

On simplifying the above relation, we get,

$48{{m}^{2}}+16>0..........\left( 6 \right)$

Now, we can observe that $48{{m}^{2}}+16$ will always be greater than zero for any real value of â€˜mâ€™. ${{m}^{2}}$will never be negative, for any real value of â€˜mâ€™. Hence, $m\in R$ is the condition for the given question that line â€˜Lâ€™ always passes through two points if it passes through focus of given parabola.

Hence, option (C) is the correct answer.

Note: One can go wrong while writing the focus of the parabola of a given equation. He/she may take (1, 0) as focus of the given parabola. But it will be wrong and need to compare with ${{\left( y-{{y}_{1}} \right)}^{2}}=4a\left( x-{{x}_{1}} \right)$ where vertex is $\left( {{x}_{1}},{{y}_{1}} \right)$and focus is $\left( {{x}_{1}}+a,{{y}_{1}} \right)$.

Hence, be careful while writing focus or vertex in these kinds of questions.

Another approach for this question would be that we can suppose two parametric coordinates on the given parabola as, \[\left( t_{1}^{2}+1,2{{t}_{1}} \right)\ and\ \left( t_{2}^{2}+1,2{{t}_{2}} \right)\]. A line is passing through (2,0) as well.

Hence, three points are collinear. Now, we know that if $\left( {{x}_{1}},{{y}_{1}} \right)$, $\left( {{x}_{2}},{{y}_{2}} \right)\ and\ \left( {{x}_{3}},{{y}_{3}} \right)$ are collinear then triangle formed by all three points have area 0. Hence,

$\left| \begin{matrix}

{{x}_{1}} & {{y}_{1}} & 1 \\

{{x}_{2}} & {{y}_{2}} & 1 \\

{{x}_{3}} & {{y}_{3}} & 1 \\

\end{matrix} \right|=0$

And slope can be given as $\dfrac{2\left( {{t}_{2}}-{{t}_{1}} \right)}{t_{2}^{2}-t_{1}^{2}}$ which is equal to m. Now, using both the equations, try to get a relation between â€˜mâ€™.

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