Question

P is a variable point on the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ whose vertex is A (a, 0). The locus of the middle point of AP is

• A. $\frac{(2x - a)^{2}}{a^{2}} - \frac{2y^{2}}{b^{2}} = 1$
• B. $\frac{(2x - a)^{2}}{a^{2}} - \frac{4y^{2}}{b^{2}} = 1$
• C. $\frac{(2x - a)^{2}}{a^{2}} - \frac{8y^{2}}{b^{2}} = 1$
• D. None of these

Answer 1

Answer: (B)

$\frac{(2x - a)^{2}}{a^{2}} - \frac{4y^{2}}{b^{2}} = 1$

Answer 2

Let (x₁, y₁) be the mid point of the variable chord AP (where A is fixed while P varies) of the hyperbola

$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$.

Equation of the chord having (x₁, y₁) as its mid points is $\frac{xx_{1}}{a^{2}} - \frac{yy_{1}}{b^{2}} = \frac{x_{1}^{2}}{a^{2}} - \frac{y_{1}^{2}}{b^{2}}$ (T = S₁)

As it passes through the fixed point A (a, 0)

∴$\frac{ax_{1}}{a^{2}} = \frac{x_{1}^{2}}{a^{2}} - \frac{y_{1}^{2}}{b^{2}}or\frac{4x_{1}^{2}}{a^{2}} - \frac{4x_{1}}{a} - \frac{4y_{1}^{2}}{b^{2}} = 0$

⇒ $\frac{4x_{1}^{2}}{a^{2}} - \frac{4x_{1}}{a} + 1 - \frac{4y_{1}^{2}}{b^{2}} = 1$

⇒ $\frac{4x_{1}^{2} - 4ax_{1} + a^{2}}{a^{2}} - \frac{4y_{1}^{2}}{b^{2}} = 1$⇒ $\frac{(2x_{1} - a)^{2}}{a^{2}} - \frac{4y_{1}^{2}}{b^{2}} = 1$

Hence locus of (x₁, y₁) is $\frac{(2x - a)^{2}}{a^{2}} - \frac{4y^{2}}{b^{2}} = 1$.