Question

Find the equation of a normal to the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=2$ at the point (4, 3).

• A. 3x + 4y = 24
• B. 4x - 3y = 7
• C. 4x + 3y = 25
• D. 3x - 4y = 0

Answer 1

The correct option is B

4x - 3y = 7

In the last topic we learned how to write the equation of tangent once the points are given.
So the tangent at (4, 3) to $\frac{x^2}{16}+\frac{y^2}{9}=2$ will be $\frac{x\times 4}{16}+\frac{y\times 3}{9}=2\Rightarrow \frac{x}{4}+\frac{y}{3}=2\text{Slope of this line}=\frac{-3}{4}\Rightarrow \text{Slope of normal}=\frac{4}{3}$
We can now write the equation of normal y - 3 = $\frac{4}{3}(x-4)$
$\Rightarrow 4x-3y=16-9=7$
Using formula
The equation of normal at $(x_1,y_1)$ to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is given by $\frac{a^{2}x}{x_1}-\frac{b^{2}y}{y_1}=a^2-b^2\Rightarrow \frac{16\times x}{4}-\frac{9\times y}{3}=16-9\Rightarrow 4x-3y=7$
We can see that using formula saves a lot time. We recommend you to remember it.