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  • A point on the ellipse,, where the normal is parallel to the l

A point on the ellipse,4x^{2}+9y^{2}=36, where the normal is parallel to the line ,4x-2y-5=0, is :

Option: 1

\left ( \frac{9}{5},\frac{8}{5} \right )              


Option: 2

\left ( \frac{8}{5},-\frac{9}{5} \right )


Option: 3

\left (- \frac{9}{5},\frac{8}{5} \right )


Option: 4

\left ( \frac{8}{5},\frac{9}{5} \right )


Answers (1)

best_answer

\\ {\text {The equation of normal of slope m to the ellipse, } \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \text { are }} \\ {y=m x \mp \frac{m\left(a^{2}-b^{2}\right)}{\sqrt{a^{2}+m^{2} b^{2}}} \text { and coordinate of point of contact is }} \\ {\left( \pm \frac{a^{2}}{\sqrt{a^{2}+m^{2} b^{2}}}, \pm \frac{m b^{2}}{\sqrt{a^{2}+m^{2} b^{2}}}\right)}

Given the equation of the ellipse is

\\\frac{x^2}{9}+\frac{y^2}{4}=1\\a^2=9,\;\;b^2=4\\\text{equation of the normal}\\y=mx+\frac{5m}{\sqrt{9+4m^2}}

This normal is parallel to the line 4x - 2y - 5 = 0.

the slope will be 2

so the equation of the normal is   

\\y=2x+\frac{10}{\sqrt{9+4(2)^2}}=2x+2\\y=2x+2

point of contact 

\\\left(\pm \frac{a^{2}}{\sqrt{a^{2}+m^{2} b^{2}}}, \pm \frac{m b^{2}}{\sqrt{a^{2}+m^{2} b^{2}}}\right)\\\left ( \pm\frac{9}{5},\pm\frac{8}{5} \right )

 

Posted by

shivangi.shekhar

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