# If one end of a focal chord of the parabola, ${{\text{y}}^2} = 16{\text{x}}$ is at (1, 4), then length of this focal chord is?

A. 25

B. 24

C. 20

D. 22

Answer

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Hint: Identify the variables in the given equation and compare it to the geometrical equation of a parabola. Upon identifying the known and required variables find the points at which both the ends of the focal chord lie, then using the formula for distance between two points determine the length of focal chord.

Complete step-by-step answer:

Given data –

Equation of parabola is ${{\text{y}}^2} = 16{\text{x}}$ and one end of the focal chord lies at (1, 4)

The geometrical equation of a parabola is in the form

${{\text{y}}^2} = 4{\text{ax}}$, whereas the ends of its focal chord are in the form $\left( {{\text{a}}{{\text{t}}^2},2{\text{at}}} \right){\text{ and }}\left( {{\text{at}}_1^2,2{\text{at}}_1^2} \right)$respectively.

And the relation between ${\text{t and }}{{\text{t}}_1}{\text{ is }}{{\text{t}}_1} = \dfrac{{ - 1}}{{\text{t}}}$.

On Comparing the equation of parabola to given equation ${{\text{y}}^2} = 16{\text{x}}$, we get

a=4.

Comparing given point (1, 4) to$\left( {{\text{a}}{{\text{t}}^2},2{\text{at}}} \right)$, we get

2at = 4

⟹at=2

We know that a=2

$ \Rightarrow {\text{t = }}\dfrac{1}{2}$

$

{{\text{t}}_1} = \dfrac{{ - 1}}{{\text{t}}} \\

\Rightarrow {{\text{t}}_1} = - 2 \\

$

Now the other end of the focal chord is $\left( {{\text{at}}_1^2,2{\text{a}}{{\text{t}}_1}} \right)$

$ \Rightarrow \left( {4{{( - 2)}^2},2(4)( - 2)} \right) = \left( {16, - 16} \right)$

Now distance between the two points (1, 4) and (16, -16) is

$

{\text{D = }}\sqrt {{{({{\text{x}}_1} - {{\text{x}}_2})}^2} + {{({{\text{y}}_1} - {{\text{y}}_2})}^2}} \\

\Rightarrow {\text{ }}\sqrt {{{(1 - 16)}^2} + {{(4 + 16)}^2}} \\

\Rightarrow \sqrt {225 + 400} \\

\Rightarrow 25. \\

$

Hence the length of the focal chord of the given parabola is 25, which makes Option A the correct answer.

Note –

In this type of question first find out and compare the equation of the given parabola. Then find you’re a, t, ${{\text{t}}_1}$ acchording to the given data in the question. Then find out the distance of the focal chord. Knowing the parabolic equation and respective properties of the focal chord is essential.

Complete step-by-step answer:

Given data –

Equation of parabola is ${{\text{y}}^2} = 16{\text{x}}$ and one end of the focal chord lies at (1, 4)

The geometrical equation of a parabola is in the form

${{\text{y}}^2} = 4{\text{ax}}$, whereas the ends of its focal chord are in the form $\left( {{\text{a}}{{\text{t}}^2},2{\text{at}}} \right){\text{ and }}\left( {{\text{at}}_1^2,2{\text{at}}_1^2} \right)$respectively.

And the relation between ${\text{t and }}{{\text{t}}_1}{\text{ is }}{{\text{t}}_1} = \dfrac{{ - 1}}{{\text{t}}}$.

On Comparing the equation of parabola to given equation ${{\text{y}}^2} = 16{\text{x}}$, we get

a=4.

Comparing given point (1, 4) to$\left( {{\text{a}}{{\text{t}}^2},2{\text{at}}} \right)$, we get

2at = 4

⟹at=2

We know that a=2

$ \Rightarrow {\text{t = }}\dfrac{1}{2}$

$

{{\text{t}}_1} = \dfrac{{ - 1}}{{\text{t}}} \\

\Rightarrow {{\text{t}}_1} = - 2 \\

$

Now the other end of the focal chord is $\left( {{\text{at}}_1^2,2{\text{a}}{{\text{t}}_1}} \right)$

$ \Rightarrow \left( {4{{( - 2)}^2},2(4)( - 2)} \right) = \left( {16, - 16} \right)$

Now distance between the two points (1, 4) and (16, -16) is

$

{\text{D = }}\sqrt {{{({{\text{x}}_1} - {{\text{x}}_2})}^2} + {{({{\text{y}}_1} - {{\text{y}}_2})}^2}} \\

\Rightarrow {\text{ }}\sqrt {{{(1 - 16)}^2} + {{(4 + 16)}^2}} \\

\Rightarrow \sqrt {225 + 400} \\

\Rightarrow 25. \\

$

Hence the length of the focal chord of the given parabola is 25, which makes Option A the correct answer.

Note –

In this type of question first find out and compare the equation of the given parabola. Then find you’re a, t, ${{\text{t}}_1}$ acchording to the given data in the question. Then find out the distance of the focal chord. Knowing the parabolic equation and respective properties of the focal chord is essential.

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