Question

The normal to the hyperbola 4x² - 9y² = 36 meets the axes in M and N and the lines MP, NP are drawn at right angles to the axes. The locus of P is the hyperbola.

• A. 9x² - 4y² = 169
• B. 4x² - 9y² = 169
• C. 3x² - 4y² = 169
• D. None of these

Answer 1

Answer: (A)

9x² - 4y² = 169

Answer 2

The given hyperbola can be written in the form

$\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$.

The equation of any normal to the given hyperbola is $\frac{3x}{\sec\theta} + \frac{2y}{\tan\theta} = 13$.

It meets x-axis in M $\left( \frac{13}{3}\sec\theta,0 \right)$ and y-axis in N $\left( 0,\frac{13}{2}\tan\theta \right)$.

Let the coordinates of P be (x₁, y₁).

Since MP ⊥ NP, x₁ = NP = $\frac{13}{2}$ secθ, y₁ = MP = $\frac{13}{3}$ tanθ

∴ secθ = $\frac{3x_{1}}{13}$ and tanθ = $\frac{2y_{1}}{13}$.

⇒ $\frac{9x_{1}^{2}}{169} - \frac{4y_{1}^{2}}{169} = 1$ ($\because$sec²θ - tan²θ = 1)

Hence the locus of (x₁, y₁) is 9x² - 4y² = 169.