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  • On the ellipse , the points at which the tangents are paral

On the ellipse \mathrm{4 x^2+9 y^2=1}, the points at which the tangents are parallel to the line 8x=9y are

Option: 1

\left(\frac{2}{5}, \frac{1}{5}\right)


Option: 2

\left(-\frac{2}{5}, \frac{1}{5}\right)


Option: 3

Both (b) and (d) 


Option: 4

\left(\frac{2}{5}, -\frac{1}{5}\right)


Answers (1)

best_answer

\mathrm{\text { Let }\left(\frac{1}{2} \cos \theta, \frac{1}{3} \sin \theta\right) \text { be a point on } 4 x^2+9 y^2=1 \text {, so }}

\mathrm{\text { equation of tangent at }\left(\frac{1}{2} \cos \theta, \frac{1}{3} \sin \theta\right) \text { is }}

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

Equating slope with 8x = 9y

\mathrm{\frac{-2 \cos \theta}{3 \sin \theta}=\frac{8}{9} \Rightarrow \tan \theta=-\frac{3}{4}}

\mathrm{\text { Hence either } \cos \theta=-\frac{4}{5}, \sin \theta=\frac{3}{5}}

\mathrm{\cos \theta=\frac{4}{5}, \sin \theta=-\frac{3}{5}}

\mathrm{\text { So the points are }\left(-\frac{2}{5}, \frac{1}{5}\right) \text { or }\left(\frac{2}{5},-\frac{1}{5}\right)}

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Ritika Harsh

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