First slide

Parabola

Question

$The straight line y = m (x - a) meets the parabola y^{2} = 4 a x in$ $two distinct points for =$

Moderate

A

$all m \in R$
B

$all m \in [- 1, 1]$
C

$all m \in R - {0}$
D

None of these

Solution

$\begin{array}{l} substitute the expression for y in the equation of the parabola \\ {(m (x - a))}^{2} = 4 a x \\ m^{2} {(x - a)}^{2} = 4 a x \\ m^{2} x^{2} - 4 a x - 2 a m^{2} x + a^{2} m^{2} = 0 \\ m^{2} x^{2} + x (- 4 a - 2 a m^{2}) + a^{2} m^{2} = 0 \\ This is a quadratic equation, if it has two roots, its discriminant must be positive \\ {(4 a + 2 a m^{2})}^{2} - 4 (m^{2}) (a^{2} m^{2}) > 0 \\ {(2 + m^{2})}^{2} - m^{4} > 0 \\ m^{4} + 4 m^{2} + 4 - m^{4} > 0 \\ 4 (m^{2} + 1) > 0 \end{array}$
$F o r a n y n o n z e r o v a l u e o f m t h i s i s p o s s i b l e$

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