Question

A chord is drawn through the focus of the parabola ${y^2} = 6x$ such that its distance from the vertex of this parabola is $\dfrac{{\sqrt 5 }}{2}$, then its slope can be :
A) $\dfrac{{\sqrt 5 }}{2}$
B) $\dfrac{{\sqrt 3 }}{2}$
C) $\dfrac{2}{{\sqrt 5 }}$
D) $\dfrac{2}{{\sqrt 3 }}$

Answer 1

Here first we will assume the slope of the chord to be m. Then we will find the equation of the chord using the focus of the parabola as passing point and then we will finally find the distance of the chord from the vertex of the parabola.
The distance of a point $\left( {{x_1},{y_1}} \right)$ to the line $ax + by + c = 0$ is given by:-
$d = \left| {\dfrac{{a\left( {{x_1}} \right) + b\left( {{y_1}} \right) + c}}{{\sqrt {{a^2} + {b^2}} }}} \right|$

Complete step-by-step answer:
Let the slope of the chord be m.
For a standard parabola:
${y^2} = 4ax$
The vertex is (0, 0) and the focus is (a, 0).
Now the given parabola is:-

$${y^2} = 6x \\\ \Rightarrow {y^2} = 4\left( {\dfrac{3}{2}} \right)x \\\$$

Comparing this equation with the standard equation we get:-
Vertex= (0, 0) and the focus is $\left( {\dfrac{3}{2},0} \right)$
Now we know that according to slope intercept form the equation of line passing through $\left( {{x_1},{y_1}} \right)$ with slope m is given by:-
$\left( {y - {y_1}} \right) = m\left( {x - {x_1}} \right)$
We know that the chord passes through the focus and has slope m.
Then according to slope intercept form the equation of the chord is:
$\left( {y - 0} \right) = m\left( {x - \dfrac{3}{2}} \right)$
Solving it further we get:-

$$y = mx - \dfrac{3}{2}m \\\ \Rightarrow mx - y - \dfrac{3}{2}m = 0 \\\$$

Now we know that the distance of a point $\left( {{x_1},{y_1}} \right)$ to the line $ax + by + c = 0$ is given by:-
$d = \left| {\dfrac{{a\left( {{x_1}} \right) + b\left( {{y_1}} \right) + c}}{{\sqrt {{a^2} + {b^2}} }}} \right|$
Hence the distance of the chord from the vertex (0, 0) is:
$d = \left| {\dfrac{{m\left( 0 \right) - 1\left( 0 \right) + \left( {\dfrac{{ - 3}}{2}m} \right)}}{{\sqrt {{{\left( m \right)}^2} + {{\left( { - 1} \right)}^2}} }}} \right|$
Solving it further we get:-
$d = \left| {\dfrac{{\dfrac{{ - 3}}{2}m}}{{\sqrt {{m^2} + 1} }}} \right|$
Now since the given distance is $\dfrac{{\sqrt 5 }}{2}$
Hence putting the value in above equation we get:-
$\dfrac{{\sqrt 5 }}{2} = \left| {\dfrac{{\dfrac{{ - 3}}{2}m}}{{\sqrt {{m^2} + 1} }}} \right|$
Simplifying it further we get:-
$\dfrac{{\sqrt 5 }}{2} = \dfrac{{3m}}{{2\sqrt {{m^2} + 1} }}$
Solving for m we get:-
$\sqrt 5 .\sqrt {{m^2} + 1} = 3m$
Squaring both the sides we get:-

$${\left( {\sqrt 5 .\sqrt {{m^2} + 1} } \right)^2} = {\left( {3m} \right)^2} \\\ \Rightarrow 5\left( {{m^2} + 1} \right) = 9{m^2} \\\$$

Solving it further we get:-

$$\Rightarrow 5{m^2} + 5 = 9{m^2} \\\ \Rightarrow 9{m^2} - 5{m^2} = 5 \\\ \Rightarrow 4{m^2} = 5 \\\$$

Solving for m we get:-
${m^2} = \dfrac{5}{4}$
Taking square root we get:-
$\sqrt {{m^2}} = \sqrt {\dfrac{5}{4}}$
Solving it further we get:-
$m = \pm \dfrac{{\sqrt 5 }}{2}$
Hence the slope of the chord is either $\dfrac{{\sqrt 5 }}{2}$ or $- \dfrac{{\sqrt 5 }}{2}$

So, the correct answer is “Option A”.

Note: Students should keep in mind that the standard equation of a parabola is ${y^2} = 4ax$ and its vertex is (0, 0) and the focus is (a, 0).
Also, students should take a note that the value of distance is always positive. So, the positive value of the distance should be considered.