Solveeit Logo
Login

Question

Question: The equation of a tangent to the parabola,\({x^2} = 8y\) which makes an angle \(\theta \)with the po...

The equation of a tangent to the parabola,x2=8y{x^2} = 8y which makes an angle θ\theta with the positive direction of x-axis, is:
a. x=ycot⁡θ+2tan⁡θ b. x=ycot⁡θ−2tan⁡θ c. x=xtan⁡θ−2cot⁡θ d. x=xtan⁡θ+2cot⁡θ  {\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 \\\

Explanation

Solution

Hint: - Tangent makes an angle θ\theta with the positive direction of x-axis , so the slope of the tangent is tan⁡θ\tan \theta . And tan⁡θ=dydx\tan \theta = \dfrac{{dy}}{{dx}}, so differentiate the equation of parabola and calculate the value of dydx\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 isx2=8y............(1){x^2} = 8y............\left( 1 \right)
Differentiation this equation w.r.t. xx
2x=8dydx ⇒dydx=x4  2x = 8\dfrac{{dy}}{{dx}} \\\ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{x}{4} \\\
Now, it is given that it makes an angle θ\theta with positive direction of x-axis.
Therefore its slope is tan⁡θ\tan \theta
And we know tan⁡θ=dydx\tan \theta = \dfrac{{dy}}{{dx}}
⇒tan⁡θ=x4 ⇒x=4tan⁡θ  \Rightarrow \tan \theta = \dfrac{x}{4} \\\ \Rightarrow x = 4\tan \theta \\\
Now from equation 1 x2=8y{x^2} = 8y
⇒8y=(4tan⁡θ)2=16tan⁡2θ ∴y=2tan⁡2θ  \Rightarrow 8y = {\left( {4\tan \theta } \right)^2} = 16{\tan ^2}\theta \\\ \therefore y = 2{\tan ^2}\theta \\\
Let, (x1,y1)\left( {{x_1},{y_1}} \right) be the points passing through the tangent.
⇒x1=4tan⁡θ, y1=2tan⁡2θ\Rightarrow {x_1} = 4\tan \theta ,{\text{ }}{y_1} = 2{\tan ^2}\theta
Equation of line passing through points (x,y)\left( {x,y} \right)and(x1,y1)\left( {{x_1},{y_1}} \right)is given as
y−y1=m(x−x1)y - {y_1} = m\left( {x - {x_1}} \right), where m is the slope
Therefore equation of tangent is
y−2tan⁡2θ=tan⁡θ(x−4tan⁡θ)y - 2{\tan ^2}\theta = \tan \theta \left( {x - 4\tan \theta } \right)
Divide by tan⁡θ\tan \theta
ytan⁡θ−2tan⁡θ=x−4tan⁡θ as, 1tan⁡θ=cot⁡θ ⇒ycot⁡θ−2tan⁡θ+4tan⁡θ=x ⇒x=ycot⁡θ+2tan⁡θ  \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. xx, then always remember that tan⁡θ=dydx\tan \theta = \dfrac{{dy}}{{dx}}, which is nothing but slope, so equate these two equations and calculate (x1,y1)\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 (x1,y1)\left( {{x_1},{y_1}} \right)points.