Question

The equation of the normal to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at the positive end of latus rectum is :

• A. $x+ey+e^{2}a=0$
• B. $x-ey-e^{3}a=0$
• C. $x-ey-e^{2}a=0$
• D. none of these

Answer 1

The correct option is B $x-ey-e^{3}a=0$

$Equationofellipse:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1Co-ordinatesofpositivelatusrectumis:(ae,\frac{b^2}{a})Equationofnormalatpoint(x1,y1)is:\frac{a^{2}x}{x1}-\frac{b^{2}y}{y1}={\left(ae\right)}^2\therefore Equationis:\frac{a^{2}x}{ae}-\frac{b^{2}y}{\frac{b^2}{a}}={\left(ae\right)}^{2}Or,\frac{x}{e}-\frac{y}{1}={ae}^{2}Or,x-ey-{ae}^3=0$

Option [B]