Question

What is the equation of the normal which is perpendicular to 3x + 4y = 5 for the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

• A. $y=\frac{4}{3}x-\frac{4(a^2+b^2)}{\sqrt{9a^2+16b^2}}$
• B. $y=\frac{4}{3}x-\frac{a^2-b^2}{\sqrt{16a^2+9b^2}}$
• C. $y=\frac{4}{3}x-\frac{4(a^2+b^2)}{\sqrt{16a^2+9b^2}}$
• D. $y=\frac{4}{3}x-\frac{4(a^2-b^2)}{\sqrt{9a^2+16b^2}}$

Answer 1

The correct option is D

$y=\frac{4}{3}x-\frac{4(a^2-b^2)}{\sqrt{9a^2+16b^2}}$

We will first find the slope of the normal, m. We know the equation of normal with slope m.
$y=mx-\frac{(a^2-b^2)m}{\sqrt{a^2+b^2{m}^2}}Slopeof3x+4y=5=\frac{-3}{4}Slopeofnormal=\frac{-1}{slopeof3x+4y=5}=\frac{\frac{-1}{-3}}{4}=\frac{4}{3}$
So the equation of normal becomes
$y=\frac{4}{3}x-\frac{(a^2-b^2)\times \frac{4}{3}}{\sqrt{a^2+b^2\times {\left(\frac{4}{3}\right)}^2}}=\frac{4}{3}x-\frac{4}{3}\frac{\frac{(a^2-b^2)}{\sqrt{9a^2+16b^2}}}{3}y=\frac{4}{3}x-\frac{4(a^2-b^2)}{\sqrt{9a^2+16b^2}}$