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  • Let the ellipse, <img alt="" src="https://lh6.googleusercontent.com/tY29nuwAWt5hNVgZsjWds7qoxml_29NfQzN-tz-6o9GSVNysXgdwn_XgCH0mvEEWqmscnoCmv8j353rzPXmNd6Jq62nDN441V6dc2KMS6-Lhegl9VWQ2Jhr9C1QWUsCt_

Let the ellipse,  pass through the point (2,3) and have eccentricity equal to . Then, the equation of the normal to the ellipse at (2,3) is:

Option: 1

2y - x = 4


Option: 2

2x - y = 1


Option: 3

3x - 2y = 0


Option: 4

3x - y =3


Answers (1)

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Equation of Normal in Slope form -

Equation of Normal in Slope form

\\ {\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)}

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