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Find the equations of the tangent and normal to the parabola y2 = 4ax at the point (at2 , 2at)
Solution:
The slope of a line is nothing but the change in y coordinate with respect to the change in x coordinate of that line.
For a curve y = f(x) containing the point (x1,y1) the equation of the tangent line to the curve at (x1,y1) is given by
y − y1 = f′(x1) (x − x1).
The equation of the given parabola is y2 = 4ax
On differentiating both sides with respect to x, we have:
2y dy/dx = 4a
⇒ dy/dx = 2a/y
Therefore, the slope of the tangent at (at2, 2at) is
dy/dx](at2, 2at) = 2a/2at = 1/t
Hence, the equation of the tangent at (at2, 2at) is
y - 2at = 1/t (x - at2)
⇒ ty - 2at2 = x - at2
⇒ ty = x + at2
Now, the slope of the normal at (at2, 2at) is
- 1/slope of the tangent at (at2, 2at)
= - 1/(1/t)
= - t
Thus, the equation of the normal at (at 2 , 2at ) is
y - 2at = - t (x - at2)
⇒ y - 2at = - tx + at3
⇒ y = - tx + 2at + at3
NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.3 Question 22
Find the equations of the tangent and normal to the parabola y2 = 4ax at the point (at2 , 2at)
Summary:
The equation of the tangent and normal to the parabola y2 = 4ax at the point (at2, 2at) is y = - tx + 2at + at3
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