Question

# A line L passing through the focus of the parabola ${{y}^{2}}=4\left( x-1 \right)$, intersects the parabola in two distinct points. If â€˜mâ€™ be the slope of the line â€˜Lâ€™ then(A) -1 < m < 1(B) m< -1 or m > 1(C) $m\in R$(D) None of the above

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.

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€™.