Equation of tangents to ellipse \frac{x^{2}}{9}+\frac{y^{2}}{4}=1 , which are perpendicular to the line 3x + 4y= -7

  • Option 1)

    4x-3y=\pm 6\sqrt{5}

  • Option 2)

    4x-3y=\pm \sqrt{12}

  • Option 3)

    4x-3y=\pm \sqrt{2}

  • Option 4)

    4x-3y=\pm {1}

 

Answers (1)

 

Standard equation -

\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}= 1
 

- wherein

a\rightarrow Semi major axis

b\rightarrow Semi minor axis

 

 

Slope of tangent =\frac{-1}{\left ( \frac{-3}{4} \right )} = \frac{4}{3}

So, equation of tangent at (3\cos \theta, 2\sin \theta )

\frac{\cos \theta x}{3}+\frac{\sin \theta y}{2} = 1

where, \frac{-\cos \theta ^{2}}{3\sin \theta } =\frac{4}{3}

\Rightarrow \tan \theta = \frac{-1}{2} \Rightarrow \sin \theta = \frac{1}{\sqrt{5}} &

                                      \cos \theta =\frac{-2}{\sqrt{5}}

\therefore equation \frac{-2}{\sqrt{5}3}x+\frac{1}{\sqrt{5}2}y=1

\Rightarrow -4x+3y=6\sqrt{5}

\Rightarrow 4x-3y=6\sqrt{5}


Option 1)

4x-3y=\pm 6\sqrt{5}

 

This solution is correct

Option 2)

4x-3y=\pm \sqrt{12}

This solution is incorrect

Option 3)

4x-3y=\pm \sqrt{2}

This solution is incorrect

Option 4)

4x-3y=\pm {1}

This solution is incorrect

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