What is the equation of the tangent at a specific point of y^2 = 4ax at (0, 0)?
(a) x = 0
(b) x = 1
(c) x = 2
(d) x = 3
I got this question during an interview for a job.
Question is from Calculus Application in division Application of Calculus of Mathematics – Class 12
NCERT Solutions for Subject Clas 12 Math Select the correct answer from above options
Interview Questions and Answers, Database Interview Questions and Answers for Freshers and Experience
Correct answer is (a) x = 0
Explanation: Equation of the given parabola is y^2 = 4ax ……….(1)
Differentiating both side of (1) with respect to x we get,
2y(dy/dx) = 4a
Or dy/dx = 2a/y
Clearly dy/dx does not exist at (0, 0). Hence, the tangent to the parabola (1) at (0, 0) is parallel to y axis.
Again, the tangent passes through (0, 0). Therefore, the required tangent to the parabola (1) at (0, 0) is the y-axis and hence the required equation of the tangent is x = 0.