JEE Main 12 Jan 2019 Evening Question 26
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
[JEE Main Online Paper Held On 12-Jan-2019 Evening]
Options:
A) $ x=y\cot \theta -2\tan \theta $
B) $ y=x\tan \theta -2\cot \theta $
C) $ x=y\cot \theta +2\tan \theta $
D) $ y=x\tan \theta +2\cot \theta $
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Answer:
Correct Answer: C
Solution:
- Given equation of parabola is $ x^{2}=8y $
$ \therefore $ $ \frac{8dy}{dx}=2x\Rightarrow \frac{dy}{dx}=\frac{x}{4}=\tan \theta $ (Given)
$ \Rightarrow x=4\tan \theta $
$ \therefore $ $ y=2{{\tan }^{2}}\theta $
Now, equation of tangent at $ (4\tan \theta ,2{{\tan }^{2}}\theta ) $ is $ y-2{{\tan }^{2}}\theta =\tan \theta $ $ (x-4\tan \theta ) $
$ \Rightarrow $ $ x=y\cot \theta +2\tan \theta $
