Question

If one end of focal chord AB of the parabola ${y^2} = 8x$ is at $A\left( {\dfrac{1}{2}, - 2} \right)$, then the equation to it at B is
1. $x + 2y + 8 = 0$
2. $2x - y - 24 = 0$
3. $x - 2y + 8 = 0$
4. $2x + y - 24 = 0$

Answer 1

Hint: In this question, we are given the equation of parabola ${y^2} = 8x$ and one end of focal chord AB i.e., $A\left( {\dfrac{1}{2}, - 2} \right)$. First step is to compare the general parabolic equation with the given one. You’ll get the value of $a$. Now, use the parametric coordinates to find the other co-ordinate of the focal chord using the condition ${t_1}{t_2} = - 1$. In last put the required co-ordinates of other end of the focal chord in the equation of tangent i.e., $y{y_1} = 2a\left( {x + {x_1}} \right)$.

Formula Used:
General equation of parabola, ${y^2} = 4ax$
Parametric co-ordinate of parabola, $\left( {a{t^2},2at} \right)$
Equation of tangent to a parabola, $y{y_1} = 2a\left( {x + {x_1}} \right)$

Complete step by step Solution:
Given that, the equation of parabola is ${y^2} = 8x - - - - - \left( 1 \right)$
General equation of parabola in this form is ${y^2} = 4ax - - - - - \left( 2 \right)$
On comparing equation (1) and (2),
It implies that, $a = 2$
Also, the parametric co-ordinates of parabola are $\left( {a{t^2},2at} \right)$
Therefore, the co-ordinates will be $A\left( {a{t_1}^2,2a{t_1}} \right)$and $B\left( {a{t_2}^2,2a{t_2}} \right)$
Where ${t_1}{t_2} = - 1 - - - - - \left( 3 \right)$
Now, one end of the focal chord is given $A\left( {\dfrac{1}{2}, - 2} \right)$
$\left( {a{t_1}^2,2a{t_1}} \right) = \left( {\dfrac{1}{2}, - 2} \right)$
$a{t_1}^2 = \dfrac{1}{2},2a{t_1} = - 2$
${t_1}^2 = \dfrac{1}{4},{t_1} = \dfrac{{ - 1}}{2}$
${t_1} = \pm \dfrac{1}{2},{t_1} = \dfrac{{ - 1}}{2}$
Here, ${t_1} \ne \dfrac{1}{2} \Rightarrow {t_1} = \dfrac{{ - 1}}{2}$
From equation (3),
${t_2} = 2$
Therefore, the co-ordinate of the other end of focal chord will be $B\left( {a{t_2}^2,2a{t_2}} \right) = \left( {8,8} \right)$
The equation of the tangent at the point $\left( {8,8} \right)$ i.e., $\left( {{x_1},{y_1}} \right)$
$ \Rightarrow y{y_1} = 2a\left( {x + {x_1}} \right)$
$8y = 2\left( 2 \right)\left( {x + 8} \right)$
$8y = 4x + 32$
$4x - 8y + 32 = 0$
$x - 2y + 8 = 0$

Hence, the correct option is 3.

Note: A parabola is a U-shaped plane curve in which any point is an equal distance from both a fixed point (also known as the focus) and a fixed straight line (known as the directrix). A parabola is a right circular cone sectioned by a plane parallel to the cone's generator.