Question

Equation of the normal to the ellipse 4(x - 1)² + 9(y - 2)² = 36, which are parallel to the line 3x - y = 1, is:

• A. 3x - y = $\sqrt{5}$
• B. 3x - y = $\sqrt{5}$ - 3
• C. 3x - y = $\sqrt{5}$ + 2
• D. 3x - y = $\sqrt{5}$ ($\sqrt{5}$ + 1)

Answer 1

Answer: (D)

3x - y = $\sqrt{5}$ ($\sqrt{5}$ + 1)

Answer 2

$\frac{\mathbf{(x - 1}\mathbf{)}^{\mathbf{2}}}{\mathbf{9}}\mathbf{+}\frac{\mathbf{(y - 2}\mathbf{)}^{\mathbf{2}}}{\mathbf{4}}\mathbf{= 1}$Equation of normal to ellipse $\frac{X^{2}}{9} + \frac{Y^{2}}{4} = 1$

3X secθ - 2Y cosecθ = 5

Slope of normal $\frac{3}{2}$tanθ = 3 ⇒ tanθ = 2

sinθ = $\frac{2}{\sqrt{5}}$, cosθ = $\frac{1}{\sqrt{5}}$

So normal is 3$\sqrt{5}$X - $\sqrt{5}$Y = 5

Now X = x - 1, Y = y - 2.

So 3$\sqrt{5}$ (x - 1) - $\sqrt{5}$ (y - 2) = 5

$\sqrt{5}$ (3x - y) = 5($\sqrt{5}$ + 1)

3x – y = $\sqrt{5}$ ($\sqrt{5}$ + 1).