Question

The equation of a tangent to the parabola,${x^2} = 8y$ which makes an angle $\theta $with the positive direction of x-axis, is:
$
  {\text{a}}{\text{. }}x = y\cot \theta + 2\tan \theta \\
  {\text{b}}{\text{. }}x = y\cot \theta - 2\tan \theta \\
  {\text{c}}{\text{. }}x = x\tan \theta - 2\cot \theta \\
  {\text{d}}{\text{. }}x = x\tan \theta + 2\cot \theta \\
$

Answer 1

Hint: - Tangent makes an angle $\theta $with the positive direction of x-axis , so the slope of the tangent is $\tan \theta $. And $\tan \theta = \dfrac{{dy}}{{dx}}$, so differentiate the equation of parabola and calculate the value of $\dfrac{{dy}}{{dx}}$. Find the touching point and use two point- formula to write the equation of the tangent.

Complete step-by-step answer:
Equation of parabola is${x^2} = 8y............\left( 1 \right)$
Differentiation this equation w.r.t. $x$
$
  2x = 8\dfrac{{dy}}{{dx}} \\
   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{x}{4} \\
$
Now, it is given that it makes an angle $\theta $with positive direction of x-axis.
Therefore its slope is $\tan \theta $
And we know $\tan \theta = \dfrac{{dy}}{{dx}}$
$
   \Rightarrow \tan \theta = \dfrac{x}{4} \\
   \Rightarrow x = 4\tan \theta \\
$
Now from equation 1 ${x^2} = 8y$
$
   \Rightarrow 8y = {\left( {4\tan \theta } \right)^2} = 16{\tan ^2}\theta \\
  \therefore y = 2{\tan ^2}\theta \\
$
Let, $\left( {{x_1},{y_1}} \right)$ be the points passing through the tangent.
$ \Rightarrow {x_1} = 4\tan \theta ,{\text{ }}{y_1} = 2{\tan ^2}\theta $
Equation of line passing through points $\left( {x,y} \right)$and$\left( {{x_1},{y_1}} \right)$is given as
$y - {y_1} = m\left( {x - {x_1}} \right)$, where m is the slope
Therefore equation of tangent is
$y - 2{\tan ^2}\theta = \tan \theta \left( {x - 4\tan \theta } \right)$
Divide by $\tan \theta $
$
  \dfrac{y}{{\tan \theta }} - 2\tan \theta = x - 4\tan \theta \\
  {\text{as, }}\dfrac{1}{{\tan \theta }} = \cot \theta \\
   \Rightarrow y\cot \theta - 2\tan \theta + 4\tan \theta = x \\
   \Rightarrow x = y\cot \theta + 2\tan \theta \\
$
Hence, option (a) is correct.

Note: - In these types of problems always remember that slope is differentiation of the equation w.r.t. $x$, then always remember that $\tan \theta = \dfrac{{dy}}{{dx}}$, which is nothing but slope, so equate these two equations and calculate $\left( {{x_1},{y_1}} \right)$ points, then from the equation of line which is stated above we can easily calculate the equation of tangent passing through $\left( {{x_1},{y_1}} \right)$points.