The line $y = m x + 1$ is a tangent to the parabola $y^{2} = 4 x$ if

m = 1

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m = 4

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m = 2

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m = 3

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Solution

The correct option is D $m = 1$
line $y = m x + 1$

$y^{2} = 4 x$

for this line to

be tangent to the parabola

${(m x + 1)}^{2} = 4 x$ ( $∴$ putting $y = m x + 1)$

$D = 0$ of

$m^{2} x^{2} + 2 m x + 1 - 4 x = 0$

$m^{2} x^{2} + (2 m - 4) x + 1 = 0$

$D = {(2 m - 4)}^{2} - 4 \times m^{2} \times 1 = 0$

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

$m^{2} - 4 m + 4 - m^{2} = 0$

$m = 1$

Suggest Corrections

Similar questions

y = m x + \frac{4 \sqrt{3}}{m}, (m \neq 0)

y^{2} = 16 \sqrt{3} x

2 x^{2} + y^{2} = 4

m^{4} + 2 m^{2} = 24

y^{2} = 16 \sqrt{3} x

2 x^{2} + y^{2} = 4

y = 2 x + 2 \sqrt{3}

y = m x + \frac{4 \sqrt{3}}{m}, (m \neq 0)

y^{2} = 16 \sqrt{3} x

2 x^{2} + y^{2} = 4

m

m^{4} + 2 m^{2} = 24

The line y = mx + 1 is a tangent to the curve y² = 4x if the value of m is

(A) 1 (B) 2 (C) 3 (D)

\frac{1}{2}

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