First slide

Parabola

Question

$Three normals are drawn to the$ $parabola y^{2} = x through point$ $(a, 0) . Then$

Moderate

A

$a = \frac{1}{2}$
B

$a = \frac{1}{4}$
C

$a > \frac{1}{2}$
D

None of the above

Solution

$\begin{array}{l} Equation of normal in slope form on y^{2} = 4 A x is \\ y = m x - 2 A m - A m^{3} - - - - - (1) \\ = m x - 2 (\frac{1}{4}) m - (\frac{1}{4}) m^{3} [\begin{array}{l} ∵ y^{2} = x \\ ∴ A = \frac{1}{4} \end{array}] \\ \Rightarrow 4 m x - 4 y - m^{3} - 2 m = 0 \\ ∵ (a, 0) lies on the normal. Then, 4 m \times a - 4 \times 0 - m^{3} - 2 m = 0 \\ \Rightarrow m (m^{2} + 2 - 4 a) = 0 \\ \Rightarrow m = 0 or m^{2} + 2 - 4 a = 0 \\ If m = 0, then from (i), \\ y = 0 i.e., x -axis is one normal. \\ If m^{2} + 2 - 4 a = 0 \Rightarrow m^{2} = 4 a - 2 [∵ m^{2} > 0] \\ \Rightarrow 4 a - 2 > 0 \Rightarrow a > \frac{1}{2} \end{array}$

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