# If y = 2x+3 is a tangent to the parabola ${y^2} = 24x$ then it’s distance from the parallel normal is given by $k\sqrt 5 $ then find the value of k.

Answer

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Hint: Here compare the equation to the standard equation of parabola then the equation of the normal and calculate the distance between the tangent and normal of parabola

Complete step-by-step answer:

y = 2x+3 is a tangent to the parabola ${y^2} = 24x$

I can also write the equation in the form of

${y^2} = 4 \times 6 \times x$ …… (1)

Eq (1) is in the form of ${y^2} = 4ax$ where a = 6

Let m be the slope of the normal

So, equation of the normal is

$y = mx - 2am - a{m^3}$

From above we have a = 6

Equation of normal $y = mx - 12m - 6{m^3}$ is parallel to y = 2x+3

As you can see y = 2x+3 is the form of y = mx+c

Where slope m = 2

On substituting m = 2 in the above equation we get,

$y = mx - 12m - 6{m^2}$

$ \Rightarrow y = 2x - 12 \times 2 - 6 \times {2^3}$

$ \Rightarrow y = 2x - 72$ …… (2)

Equation (2) is the required equation of the normal

Distance between y = 2x+3 and y = 2x-72 is

$d = \left| {\dfrac{{3 - \left( { - 72} \right)}}{{\sqrt {{2^2} + 1} }}} \right|$

$d = \left| {\dfrac{{75}}{{\sqrt 5 }}} \right|$

On rationalizing the denominator

$d = \dfrac{{75}}{{\sqrt 5 }} \times \dfrac{{\sqrt 5 }}{{\sqrt 5 }}$

$d = \dfrac{{75\sqrt 5 }}{5}$

$d = 15\sqrt 5 $

On comparing this by $k\sqrt 5 $ we get the value of k as 15

NOTE:

Whenever you come across this type of problem first compare the equation to the standard equation of parabola . After that get the equation of the normal and calculate the distance between the tangent and normal of the parabola. On comparing this answer to the given answer we get the final answer.

Complete step-by-step answer:

y = 2x+3 is a tangent to the parabola ${y^2} = 24x$

I can also write the equation in the form of

${y^2} = 4 \times 6 \times x$ …… (1)

Eq (1) is in the form of ${y^2} = 4ax$ where a = 6

Let m be the slope of the normal

So, equation of the normal is

$y = mx - 2am - a{m^3}$

From above we have a = 6

Equation of normal $y = mx - 12m - 6{m^3}$ is parallel to y = 2x+3

As you can see y = 2x+3 is the form of y = mx+c

Where slope m = 2

On substituting m = 2 in the above equation we get,

$y = mx - 12m - 6{m^2}$

$ \Rightarrow y = 2x - 12 \times 2 - 6 \times {2^3}$

$ \Rightarrow y = 2x - 72$ …… (2)

Equation (2) is the required equation of the normal

Distance between y = 2x+3 and y = 2x-72 is

$d = \left| {\dfrac{{3 - \left( { - 72} \right)}}{{\sqrt {{2^2} + 1} }}} \right|$

$d = \left| {\dfrac{{75}}{{\sqrt 5 }}} \right|$

On rationalizing the denominator

$d = \dfrac{{75}}{{\sqrt 5 }} \times \dfrac{{\sqrt 5 }}{{\sqrt 5 }}$

$d = \dfrac{{75\sqrt 5 }}{5}$

$d = 15\sqrt 5 $

On comparing this by $k\sqrt 5 $ we get the value of k as 15

NOTE:

Whenever you come across this type of problem first compare the equation to the standard equation of parabola . After that get the equation of the normal and calculate the distance between the tangent and normal of the parabola. On comparing this answer to the given answer we get the final answer.

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