The tangent to an ellipse x2 + 16y2 = 16 and making angle 60° with x-axis is:

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  1. x - √3y + 7 = 0
  2. ​​x + √3y - 7 = 0
  3. ​√3x - y + 7 = 0
  4. √3x + y + 7 = 0

Answer (Detailed Solution Below)

Option 3 : ​√3x - y + 7 = 0
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NIMCET 2020 Official Paper
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120 Questions 480 Marks 120 Mins

Detailed Solution

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Concept:

Straight Lines:

  • The general equation of a line is y = mx + c, where m is the slope of the line.
  • If the line y = mx + c makes an angle θ with the positive direction of the x-axis, then m = tan θ.

 

Tangents to an Ellipse:

  • If the line y = mx + c touches the ellipse \(\rm \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\), then c2 = a2m2 + b2.
  • The straight lines \(\rm y = mx \pm \sqrt{a^2m^2 + b^2}\) represent the tangents to the ellipse.

 

Calculation:

The given tangent makes an angle of 60 with the positive direction of the x-axis.

∴ m = tan 60 = √3.

Converting the given equation of the ellipse into the standard form \(\rm \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 \):

x2 + 16y2 = 16

⇒ \(\rm \dfrac{x^2}{4^2}+\dfrac{y^2}{1^2}=1 \)

∴ a = 4 and b = 1.

Using the formula \(\rm y = mx \pm \sqrt{a^2m^2 + b^2} \) for the tangents to the ellipse, the required equation is:

\(\rm y = \sqrt3x + \sqrt{4^2({\sqrt3})^2 + 1^2} \)

⇒ y = √3x + 7

⇒ ​√3x - y + 7 = 0.

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