First slide

Parabola

Question

$If y = 2 x - 3 is a tangent to the parabola y^{2} = 4 a (x - \frac{1}{3}), then a is equal to$

Moderate

A

$\frac{22}{3}$
B

$- 1$
C

$\frac{14}{3}$
D

$- \frac{14}{3}$

Solution

$Solving y = 2 x - 3$
$and y^{2} = 4 a (x - \frac{1}{3})$
we have
$(2 x - 3)^{2} = 4 a (x - \frac{1}{3})$
$\begin{aligned} or 4 x^{2} + 9 - 12 x = 4 a x - \frac{4 a}{3} \\ or 4 x^{2} - 4 (3 + a) x + 9 + \frac{4 a}{3} = 0 \end{aligned}$
This equation must have equal roots, i.e., D = 0
$or 16 (3 + a)^{2} - 16 (9 + \frac{4 a}{3}) = 0$
$\begin{array}{l} or 9 + a^{2} + 6 a = 9 + \frac{4 a}{3} \\ or a^{2} + \frac{14 a}{3} = 0 \\ or a = 0 or a = - \frac{14}{3} \end{array}$

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