Question

If (2,−8) is one end of focal chord of the parabola ${{\text{y}}^2} = 32{\text{x}}$, then the other end of the focal chord, is
A. (32,32)
B. (32, −32)
C. (−2,8)
D. (2,8)

Answer 1

Hint- Proceed the solution of this question first by comparing it with general parabola hence find the focal point then consider a general point on the parabola in terms of t then find the value of t with known coordinates hence t is known we can find other end coordinates too. Use the relation between parametric coordinates of the focal chord that is $\Rightarrow {{\text{t}}_1}{{\text{t}}_2} = - 1$.

Complete step-by-step answer:

In this particular question it is given one end of focal chord of the parabola ${{\text{y}}^2} = 32{\text{x}}$
So on comparing it with general parabola equation ${{\text{y}}^2} = 4.{\text{a}}{\text{.x}}$
$\Rightarrow 32{\text{x = 4}}{\text{.a}}{\text{.x}}$
On solving
$\Rightarrow {\text{a = 8}}$
As we can assume a general point $\left( {{\text{a}}{{\text{t}}^2},2{\text{at}}} \right)$ on the parabola ${{\text{y}}^2} = 4.{\text{a}}{\text{.x}}$
Vertex of parabola ≡ (0,0)

Hence we have to take two points on the parabola as extremities of the chord that we can assume as \left( {{\text{at}}_1^2,2{\text{a}}{{\text{t}}_1}} \right){\text{ & }}\left( {{\text{at}}_2^2,2{\text{a}}{{\text{t}}_2}} \right)
If the chord of the parabola which passes through the focus, then it is called the focal chord.

If $y^2 = 4ax$ be the equation of a parabola and $\left( {{\text{at}}_1^2,2{\text{a}}{{\text{t}}_1}} \right)$ a point P on it. Suppose the coordinates of the other extremity Q of the focal chord through P are $\left( {{\text{at}}_2^2,2{\text{a}}{{\text{t}}_2}} \right)$
Then, PS and SQ, where S is the focus (a, 0), have the same slopes
so on equalising the slope of both line PS and SQ, we will get a condition of focal chord i.e.
$\Rightarrow {{\text{t}}_1}{{\text{t}}_2} = - 1$ ………… (1)
Now we have a relation between {{\text{t}}_1}{\text{ & }}{{\text{t}}_2} so we can write coordinates of ${{\text{t}}_2}{\text{ in terms of }}{{\text{t}}_1}{\text{ }}$which will be
\Rightarrow {\text{at}}_2^2 = {\text{ a}}{\left( {\dfrac{1}{{{{\text{t}}_1}}}} \right)^2}{\text{ & }}2{\text{a}}{{\text{t}}_2} = 2{\text{a}}\left( {\dfrac{{ - 1}}{{{{\text{t}}_1}}}} \right) ……. (2)
So the coordinates of one end of focal chord of the parabola ${{\text{y}}^2} = 32{\text{x}}$ given in the question which is (2,-8) so on comparing these with general coordinate $\left( {{\text{at}}_1^2,2{\text{a}}{{\text{t}}_1}} \right)$
$\Rightarrow {\text{at}}_1^2 = 2,\& {\text{ }}2{\text{a}}{{\text{t}}_1} = - 8$
on putting a= 8
$\Rightarrow 2.8{{\text{t}}_1} = - 8$
$\Rightarrow {{\text{t}}_1} = \dfrac{{ - 1}}{2}$
Hence on putting value of ${{\text{t}}_1}$ and ${\text{a}}$ in equation (2) we can find the coordinates of other end of the focal chord
Therefore,
$\Rightarrow {\text{at}}_2^2 = {\text{ a}}{\left( {\dfrac{1}{{{{\text{t}}_1}}}} \right)^2}{\text{ = 8}} \times {\left( { - 2} \right)^2}{\text{ = 32}}$
${\text{ }}2{\text{a}}{{\text{t}}_2} = 2{\text{a}}\left( {\dfrac{{ - 1}}{{\dfrac{{ - 1}}{2}}}} \right) = 2 \times 8 \times 2 = 32$
Therefore, coordinates of other end of the focal chord will be (32, 32)
Hence, option A is correct.

Note- There is also an another approach to solve such type of particular question-
There is result , that we should have remember that, if \left( {{{\text{x}}_1},{{\text{y}}_1}} \right){\text{ & }}\left( {{{\text{x}}_2},{{\text{y}}_2}} \right) are two ends of focal chord of parabola ${{\text{y}}^2} = 32{\text{x}}$, then
$\Rightarrow {{\text{x}}_1}.{{\text{x}}_2} = {{\text{a}}^2}$
Given equation of parabola-
$\Rightarrow {{\text{y}}^2} = 32{\text{x}}$
Vertex of parabola ≡ (0,0)
∴a=8
One end of the chord = (2, −8)
Let the other end of the chord be (x, y).
∴$2.{\text{x}} = {8^2}$
⇒x=32
Substituting the value of x in the given equation of the parabola, we have
${{\text{y}}^2} = 32.32$
y = 32
As the one end of the focal chord is on (2, −8) hence the other end of the chord will be (32, 32).

Hence the focal cord and its extremities shown in below figure.