Question

Find the condition for the line $y = mx + c$ to be a tangent to the parabola ${x^2} = 4ay$

Answer 1

Hint: We will differentiate the equation of the parabola. Then we will take a point $\left( {{x_0},{y_0}} \right)$ as a point on the parabola. Then we will equate slope m of the given line to the value of $\dfrac{{dy}}{{dx}}$ at the point $\left( {{x_0},{y_0}} \right)$ . Then we will obtain ${x_0}$ in terms of m and by substituting it in the equation of parabola we can obtain ${y_0}$ in terms of m. Then we can find the value of c in terms of m by substituting the values of ${x_0}$ and ${y_0}$ in the equation of the straight line to obtain the required condition.

Complete step-by-step answer:
We have the equation of the parabola as ${x^2} = 4ay$
We can write it as y equals to some function of x.
 $ \Rightarrow y = \dfrac{{{x^2}}}{{4a}}$
Now we can differentiate it with respect to x.
 $ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\dfrac{{{x^2}}}{{4a}}} \right)$
We can take the constant term outside,
 $ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{4a}}\dfrac{d}{{dx}}{x^2}$
We know that $\dfrac{d}{{dx}}\left( {{x^2}} \right) = 2x$
 $ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{2x}}{{4a}}$
 $ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{x}{{2a}}$
We know that slope of the tangent to a curve at the point $\left( {{x_0},{y_0}} \right)$ is given by ${\left. {\dfrac{{dy}}{{dx}}} \right|_{\left( {{x_0},{y_0}} \right)}}$
So, the slope of the tangent to the parabola at $\left( {{x_0},{y_0}} \right)$ is given by,
 $ \Rightarrow {\left. {\dfrac{{dy}}{{dx}}} \right|_{\left( {{x_0},{y_0}} \right)}} = \dfrac{{{x_0}}}{{2a}}$
Now we can equate this slope with the slope m of the equation of straight line $y = mx + c$ .
 $ \Rightarrow m = \dfrac{{{x_0}}}{{2a}}$
On rearranging, we get,
 $ \Rightarrow {x_0} = 2am$
Now we can find ${y_0}$ by substituting in the equation of the parabola.
 $ \Rightarrow {y_0} = \dfrac{{{x_0}^2}}{{4a}}$
 $ \Rightarrow {y_0} = \dfrac{{{{\left( {2am} \right)}^2}}}{{4a}}$
 $ \Rightarrow {y_0} = \dfrac{{4{a^2}{m^2}}}{{4a}}$
 $ \Rightarrow {y_0} = a{m^2}$
Now we substitute the values of ${x_0}$ and ${y_0}$ in the equation of the straight line.
 $ \Rightarrow {y_0} = m{x_0} + c$
 $ \Rightarrow a{m^2} = m\left( {2am} \right) + c$
 $ \Rightarrow a{m^2} = 2a{m^2} + c$
 $ \Rightarrow c = - a{m^2}$
So, the condition for the line $y = mx + c$ to be a tangent to the parabola ${x^2} = 4ay$ is when $c = - a{m^2}$ .
We can write the equation of the tangent as $y = mx - a{m^2}$ .

Note: The derivative of a curve at a point will give the slope of the tangent to the curve at that particular point. A parabola is the collection of all the points which is equidistant from a fixed point and a straight line. The equation $y = mx + c$ is the slope intercept form of the equation of straight line where m is the slope of the line and c is the y coordinate of point at which the line crosses the y axis. We must not directly substitute for slope using the derivative in the slope intercept form. We must assume an arbitrary point on the parabola.