Question

Slope of the line touching both parabolas \[{y^2} = 4ax\] and \[{x^2} = - 32y\] is
A.\[\dfrac{1}{2}\]
B.\[\dfrac{3}{2}\]
C.\[\dfrac{1}{8}\]
D.\[\dfrac{2}{3}\]

Answer 1

Hint: Given are the equation of the parabolas. We know the equation given the tangent equation since line is common to both the parabolas, we can equate them to get the value of the slope. Also the line touching is at same point so can be considered as tangent

Complete step by step solution:
Given that, \[{y^2} = 4ax\] and \[{x^2} = - 32y\] are the two given parabolas. So we need to find the slope of the line that touches both of these.
First we will go for, \[{y^2} = 4ax\]
Equation of tangent is given by the equation \[y = mx + \dfrac{a}{m}\]
If a is equals to 1 then we can write,
\[y = mx + \dfrac{1}{m}\] …..equation1
Now for the second parabola,
\[{x^2} = - 32y\]
Equation of tangent is given by the equation \[y = mx - a{m^2}\]
Here \[4a = 32\]
Thus the value of a will be \[a = 8\]
Then the equation of the tangent will become,
\[y = mx + 8{m^2}\] …….equation2
Now on equation the slope part we get,
\[8{m^2} = \dfrac{1}{m}\]
On taking m on one side,
\[{m^3} = \dfrac{1}{8}\]
Taking the cube root we get,
\[m = \dfrac{1}{2}\]
This is the slope of the line touching the parabolas \[m = \dfrac{1}{2}\]
So, the correct answer is “\[m = \dfrac{1}{2}\]”.

Note: Here note that since the same line is touching both the parabolas. So we can equate the slope of the tangent otherwise we cannot directly equate the slopes like this. Both the parabolas are not the same facing. We need not to find the equation of tangent as in because only slope is asked. And if required we can use any of the two equations.