First slide

Parabola

Question

If y = mx + 4 is a tangent to both the parabolas $y^{2} = 4 x$ and $x^{2} = 2 b y$ , then b is equal to

Moderate

A

-64
B

-32
C

-128
D

128

Solution

The given line equation is
y = mx+4                       ….(i)
The equation of the tangent to the parabola $y^{2} = 4 x$ is    $y = m x + \frac{1}{m}$                         ….(ii)
from (i) and (ii)
$4 = \frac{1}{m} \Rightarrow m = \frac{1}{4}$
So line $y = \frac{1}{4} x + 4$ is also tangent to parabola $x^{2} = 2 b y$ , so solve
$x^{2} = 2 b (\frac{x + 16}{4})$ , it must have only one solution, it implies that discriminant must be zero.
$\Rightarrow 2 x^{2} - b x - 16 b = 0 \Rightarrow D = 0 \Rightarrow b^{2} - 4 \times 2 \times (- 16 b) = 0$
$\Rightarrow b^{2} + 32 \times 4 b = 0$
$b = - 128, b = 0$ (not possible)

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