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NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Ellipse

NTA Abhyas 2020 · Mathematics · Ellipse

Two lines L _ 1 and L _ 2 of slope 1 are tangent to y^ 2 = 4x and...

Two lines

{L}_{1}

and

{L}_{2}

of slope

1

are tangent to

y^{2} = 4x

and

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

respectively, such that the distance

d

units between

{L}_{1}

and

{L}_{2}

is minimum, then the value of

d

is equal to

Options

A $\sqrt{3} - \dfrac{1}{\sqrt{2}}$
B $\sqrt{3} + \dfrac{1}{\sqrt{2}}$
C $\sqrt{2} - 1$
D $\sqrt{3} + \sqrt{2}$

Correct answer

A. $\sqrt{3} - \dfrac{1}{\sqrt{2}}$

Step-by-step solution

y = x + 1

is a tangent to

y^{2} = 4x

y = x \pm \sqrt{4 + 2}

is a tangent to

\dfrac{x^{2}}{4} + \dfrac{y^{2}}{2} = 1

⇒

Minimum distance will be between the parallel lines

y = x + \sqrt{6}

and

y = x + 1

⇒ d = \dfrac{\sqrt{6} - 1}{2} = \sqrt{3} - \dfrac{1}{2}

& units

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