If $y = m x + c$ touches the parabola $y^{2} = 4 a (x + a)$ , then

c = \frac{a}{m}

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c = a m + \frac{a}{m}

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c = a + \frac{a}{m}

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Solution

The correct option is A $c = a m + \frac{a}{m}$
Given equation of line
$y = m x + c$

Given the equation of a parabola

$y^{2} = 4 a (x + a)$

$\Rightarrow (m x + c)^{2} = 4 a x + 4 a^{2}$

$\Rightarrow m^{2} x^{2} + (2 m c - 4 a) x + (c^{2} - 4 a^{2}) = 0$

Since line touches the parabola i.e. there is only one solution.

$\Rightarrow D = 0$

$\Rightarrow (2 m c - 4 a)^{2} - 4 m^{2} (c^{2} - 4 a^{2}) = 0$

$\Rightarrow m c = a (1 + m^{2})$

$\Rightarrow c = \frac{a (1 + m^{2})}{m}$

$\Rightarrow c = \frac{a}{m} + a m$

Suggest Corrections

Similar questions

y = m x + c

y^{2} = 4 a (x + a)

y = m x + c

y^{2} = 4 a (x + a)

a m + \frac{a}{m}

y^{2} = 4 a (x + a)

c = m a + \frac{a}{m} .

y = m x + c

y^{2} = 4 a (x + a)

y = m (x + a) + \frac{a}{m}

y^{2} = 4 a (x + a)

m

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