Question

The equation of the normal at the point (6,4) on the hyperbola $\frac{x^2}{9}-\frac{y^2}{16}=3$ is

• A. 3x + 8y = 50
• B. 3x - 8y = 50
• C. 8x + 3y = 50
• D. 8x - 3y = 50

Answer 1

The correct option is A

3x + 8y = 50

Like we have done before, solving this is only a matter of finding the slope of the normal and using the slope-point form of arriving at the equation of a line.

$\frac{dy}{dx{|}_{(6,4)}}=\frac{8}{3}$

Slope of normal =$-\frac{1}{slopeoftangent}$

$=-\frac{1}{\frac{dy}{dx{|}_{(6,4)}}}$

$=-\frac{3}{8}$

$\therefore$ Required equation of normal is,

$y-4=-\frac{3}{8}(x-6)$

$8y-32=-3x+18$

$3x+8y=50$ is the required equation of normal