Question

The equation of the normal to the curve $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at the point $(x_1,y_1)$ on it is?

• A. $\frac{a^{2}x}{x_1}-\frac{b^{2}y}{y_1}=a^2-b^2$
• B. $\frac{a^x}{x_1}+\frac{b^{2}y}{y^1}=a^2+b^2$
• C. $a^{2}x{x}_1+b^{2}y{y}_1=1$
• D. $b^{2}x{x}_1-a^{2}y{y}_1=0$

Answer 1

The correct option is A $\frac{a^{2}x}{x_1}-\frac{b^{2}y}{y_1}=a^2-b^2$

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

$\frac{2x}{a^2}\frac{dx}{dy}+\frac{2y}{b^2}=0$

$-\frac{dx}{dy}=\frac{a^2{y}_1}{b^2{x}_1}$

Line - $y=mx+c$

$y=\frac{a^2{y}_1}{b^2{x}_1}x+c$

Putting $(x_1,y_1)$ in the line equation,we get

$y_1=\frac{a^2{y}_1}{b^2}+c$

$c=y_1(1-\frac{a^2}{b^2})$

$\frac{y}{y_1}=\frac{a^{2}x}{b^2}+\frac{b^2-a^2}{b^2}$

$\frac{a^{2}x}{x_1}-\frac{b^{2}y}{y_1}=a^2-b^2$