# Equation of a common tangent to the curves \[{y^2} = 8x\]and \[xy = - 1\] is

\[

{\text{A}}{\text{. }}3y = 9x + 2 \\

{\text{B}}{\text{. }}y = 2x + 1 \\

{\text{C}}{\text{. }}2y = x + 8 \\

{\text{D}}{\text{. }}y = x + 2 \\

\]

Last updated date: 23rd Mar 2023

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Answer

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Hint:-Here, we write the equation of tangent to the parabola in slope form and then find the value of $m$ to get the equation of tangent.

Given equation of parabola \[{y^2} = 8x\]

And general form of parabola is \[{y^2} = 4ax\]

By comparing the equation we get \[a = 2\]

Equation of a tangent to parabola \[{y^2} = 8x\] is

We know that equation of tangent of parabola in the form of slope is \[y = mx + \frac{a}{m}{\text{ }}\]

Put the value of $a$ in a tangent equation we get \[y = mx + \frac{2}{m}{\text{ }}......{\text{(i) }}\]

Now solving $({\text{i)}}$with \[xy = - 1\]

\[

x\left( {mx + \frac{2}{m}} \right) = - 1 \\

\Rightarrow m{x^2} + \left( {\frac{2}{m}} \right)x + 1 = 0 \\

\]

Now for the tangent to the discriminant of the above quadratic should be zero because we know for the tangent it must touch at point only. By making discriminant equal to zero we only get one point. And this is the equation of tangent.

\[

{\left( {\frac{2}{m}} \right)^2} - 4m = 0 \\

\Rightarrow 4 - 4{m^3} = 0 \\

\Rightarrow {m^3} - 1 = 0 \\

\Rightarrow {m^3} = 1{\text{ }} \\

\]

\[\therefore m = 1\] Only real solution

Now put the value of $m$ in equation $({\text{i)}}$

Hence required common tangent is \[y = x + 2\]

Here option D is the correct answer.

Note: - Whenever we face such a type of question we have to assume the tangent equation in slope form and equate it with the given equation to find the value of slope that we assumed in the tangent equation. And then by putting the value of slope we get the required tangent.

Given equation of parabola \[{y^2} = 8x\]

And general form of parabola is \[{y^2} = 4ax\]

By comparing the equation we get \[a = 2\]

Equation of a tangent to parabola \[{y^2} = 8x\] is

We know that equation of tangent of parabola in the form of slope is \[y = mx + \frac{a}{m}{\text{ }}\]

Put the value of $a$ in a tangent equation we get \[y = mx + \frac{2}{m}{\text{ }}......{\text{(i) }}\]

Now solving $({\text{i)}}$with \[xy = - 1\]

\[

x\left( {mx + \frac{2}{m}} \right) = - 1 \\

\Rightarrow m{x^2} + \left( {\frac{2}{m}} \right)x + 1 = 0 \\

\]

Now for the tangent to the discriminant of the above quadratic should be zero because we know for the tangent it must touch at point only. By making discriminant equal to zero we only get one point. And this is the equation of tangent.

\[

{\left( {\frac{2}{m}} \right)^2} - 4m = 0 \\

\Rightarrow 4 - 4{m^3} = 0 \\

\Rightarrow {m^3} - 1 = 0 \\

\Rightarrow {m^3} = 1{\text{ }} \\

\]

\[\therefore m = 1\] Only real solution

Now put the value of $m$ in equation $({\text{i)}}$

Hence required common tangent is \[y = x + 2\]

Here option D is the correct answer.

Note: - Whenever we face such a type of question we have to assume the tangent equation in slope form and equate it with the given equation to find the value of slope that we assumed in the tangent equation. And then by putting the value of slope we get the required tangent.

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