Find the equations of common tangents to the parabolas $y^{2} = 16 x$ and the circle $x^{2} + y^{2} = 8$

y = (x + 4)

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y = - (x - 4)

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y = - (x + 4)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

y = (x - 4)

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Solution

The correct options are
A $y = (x + 4)$
C $y = - (x + 4)$
$y^{2} = 16 x ∴ 4 a = 16$ or $a = 4.$
Any tangent to parabola is
$y = m x + \frac{a}{m}$ or $m x - y + \frac{4}{m} = 0.$
Apply $p = r$ with the circle $(0, 0), r = \sqrt{8}$ for tangency.
$∴ \frac{4 / m}{\sqrt{m^{2} + 1}} = \sqrt{8}$
or $2 = m^{4} + m^{2}$ or $(m^{4} + m^{2} - 2) = 0$
or $(m^{2} + 2) (m^{2} - 1) = 0 ∴ m = \pm 1$
Hence the common tangents are $y = \pm (x + 4) .$
Ans: A,C

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Equation of the common tangent to the parabola $y^{2} = 24 x$ and the circle $x^{2} + y^{2} = 18$ is

y^{2} = 16 x

x y = - 4

Find the equation of normal to the parabola
$y^{2} = 16 x$ at point (4,8)

y^{2} = 8 x

x^{2} + y^{2} - 12 x + 4 = 0

y^{2} = 24 x

x^{2} + y^{2} = 18

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