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Parabola

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Question

The equation of the line touching both the parabola y² = 4x and $x^{2} = - 32 y is ax + by + c = 0$ . Then the value of a + b +c is

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Solution

The equation of tangent in terms of the slope of the parabola y² = 4x is $y = mx + \frac{1}{m}$
Since (i) is also the tangent of x² = - 32 !, wa have
$\begin{aligned} x^{2} = - 32 (mx + \frac{1}{m}) \\ or x^{2} + 32 mx + \frac{32}{m} = 0 \end{aligned}$
The above equation must have equal roots.
Hence, its discriminant must be zero. Therefore,
$\begin{aligned} (32 m)^{2} = 4 \times 1 \times \frac{32}{m} \\ i . e ., m^{3} = \frac{1}{8} or m = \frac{1}{2} \end{aligned}$
From (i), $y = \frac{x}{2} + 2 or x - 2 y + 4 = 0$

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