Question

If $y = 4x + d$ is a tangent to the parabola \[{y^2} = 4\left( {x + 1} \right)\] then 4d is
(A). 9
(B). 24
(C). 17
(D). 2

Answer 1

Hint: In order to solve the problem first find the general slope of the tangent to the given parabola, then use the slope of the given tangent in the problem to find the point of intersection of the parabola and the tangent. Finally use the point in the equation of the line to find the value of d.

Complete step-by-step answer:

Given equation of the tangent: $y = 4x + d$

Given equation of the parabola: \[{y^2} = 4\left( {x + 1} \right)\]

We know that the slope of tangent to any curve is given by differentiating the curve with respect to x variable.

In order to find the general slope of the given parabola let us differentiate the equation of parabola with respect to x.

Slope of tangent:

$

   \Rightarrow \dfrac{d}{{dx}}\left[ {{y^2}} \right] = \dfrac{d}{{dx}}\left[ {4\left( {x + 1} \right)} \right] \\

   \Rightarrow \dfrac{d}{{dx}}\left[ {{y^2}} \right] = \dfrac{{d\left( {4x} \right)}}{{dx}} + \dfrac{{d\left( 4 \right)}}{{dx}} \\

   \Rightarrow 2y\dfrac{{dy}}{{dx}} = 4 + 0{\text{ }}\left[ {\because \dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\& \dfrac{d}{{dx}}\left( k \right) = 0} \right] \\

   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{4}{{2y}} \\

   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{2}{y} \\

 $

Now let us compare the slope of the general tangent to the slope of the given tangent in order to find the point of intersection.

Let the point of intersection of the tangent and the parabola be at point $P\left( {{x_1},{y_1}} \right)$

So for the given point P the slope of the tangent will be

Slope of tangent:

$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{2}{{{y_1}}}$

But for the given tangent equation which is $y = 4x + d$ of the form $y = mx + c$ slope of the given tangent is $4$.

In order to find the point of intersection let us compare both the slopes.

\[

   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{2}{{{y_1}}} = 4 \\

   \Rightarrow \dfrac{2}{{{y_1}}} = 4 \\

   \Rightarrow {y_1} = \dfrac{2}{4} \\

   \Rightarrow {y_1} = \dfrac{1}{2} \\

 \]

Now let us put the value of ${y_1}$ in the equation of the parabola to find the value of \[{x_1}\]

$

   \Rightarrow {\left( {{y_1}} \right)^2} = 4\left( {{x_1} + 1} \right) \\

   \Rightarrow {\left( {\dfrac{1}{2}} \right)^2} = 4\left( {{x_1} + 1} \right) \\

   \Rightarrow 4{x_1} + 4 = \dfrac{1}{4} \\

   \Rightarrow 4{x_1} = \dfrac{1}{4} - 4 \\

   \Rightarrow 4{x_1} = \dfrac{{1 - 16}}{4} \\

   \Rightarrow 4{x_1} = \dfrac{{15}}{4} \\

   \Rightarrow {x_1} = \dfrac{{15}}{{16}} \\

 $

So, the point of intersection of parabola and the tangent is $P\left( {{x_1},{y_1}} \right) =

P\left( {\dfrac{{ - 15}}{{16}},\dfrac{1}{2}} \right)$

As this point $P\left( {{x_1},{y_1}} \right) = P\left( {\dfrac{{ - 15}}{{16}},\dfrac{1}{2}} \right)$

lie on the tangent. So it must satisfy the equation of the tangent.

Let us put the value in the equation of tangent to find the value of d.

\[

   \Rightarrow {y_1} = 4{x_1} + d \\

   \Rightarrow \dfrac{1}{2} = 4 \times \dfrac{{ - 15}}{{16}} + d \\

   \Rightarrow \dfrac{1}{2} = \dfrac{{ - 15}}{4} + d \\

   \Rightarrow d = \dfrac{1}{2} + \dfrac{{15}}{4} \\

   \Rightarrow d = \dfrac{{2 + 15}}{4} \\

   \Rightarrow d = \dfrac{{17}}{4} \\

 \]

Now let us find the value of 4d.

\[

  \because d = \dfrac{{17}}{4} \\

   \Rightarrow 4d = 17 \\

 \]

Hence, the value of 4d is 17.

So, option R is the correct option.

Note: In order to solve such types of problems students must remember the general equation of different types of curves like line, parabola etc. Also the students must remember that the general slope of the tangent to any figure is given by differentiating the equation of curve with respect to variable x.