Question

The feet of the normals to $y^2 = 4ax$ from the point $(6a, 0)$ are:

  • $(0, 0)$

  • $(4a, -4a)$

  • $(4a, -4a)$

  • $(0, 0), (4a, 4a), (4a, -4a)$

✨ Free AI Tools for You

Powered by Chatterbot AI

Answer

To find the feet of the normals to the parabola given by ( y^2 = 4ax ) from the point ( (6a, 0) ), we need to follow a systematic approach involving the properties of the parabola and the equations of the normals.

Steps to Find the Feet of the Normals:

  1. Equation of the Parabola and its Derivation:

    • The given parabola is ( y^2 = 4ax ).

    • The parametric form for this parabola is ( (at^2, 2at) ).

  2. Normal to the Parabola:

    • For the point ((at^2, 2at)), the equation of the normal is: $$ y + tx = 2at + at^3 $$ Plugging in the coordinates of the given point ((6a, 0)), we get: $$ 0 + t \cdot 6a = 2at + at^3 $$

    • Simplifying, we obtain: $$ 6at = 2at + at^3 \quad \Rightarrow \quad 4at = at^3 \quad \Rightarrow \quad t(t^2 - 4) = 0 $$

  3. Solving for ( t ):

    • The equation ( t(t^2 - 4) = 0 ) gives us three solutions: ( t = 0 ), ( t = 2 ), and ( t = -2 ).

  4. Finding the Feet of the Normals:

    • For ( t = 0 ): [ (at^2, 2at) = (0, 0) ]

    • For ( t = 2 ): [ (at^2, 2at) = (4a, 4a) ]

    • For ( t = -2 ): [ (at^2, 2at) = (4a, -4a) ]

The feet of the normals to the parabola ( y^2 = 4ax ) from the point ((6a, 0)) are ( (0, 0) ), ( (4a, 4a) ), and ( (4a, -4a) ).

Thus, the correct answer is: (0, 0), (4a, 4a), (4a, -4a).


Was this helpful?

India's 1st AI Doubt Solver for CBSE, JEE, and NEET

Ask a Question for Free

and then it's just ₹212 a month