Solveeit Logo
Login

Question

Question: If the normal at P to the rectangular hyperbola x<sup>2</sup> – y<sup>2</sup> = 4 meets the axis in ...

If the normal at P to the rectangular hyperbola x2 – y2 = 4 meets the axis in G and g and C is the centre of the hyperbola, then-

A

2PG = PC

B

12\frac{1}{2}Pg = PC

C

2\sqrt{2}PG = Pg

D

Gg = 2PC

Answer

Gg = 2PC

Explanation

Solution

Let P(x1, y1) be any point on the hyperbola

x2 – y2 = 4, then equation of the normal at P is y – y1 = –y1x1\frac{y_{1}}{x_{1}} (x – x1) Ž x1y + y1x = 2x1y1.

Then coordinates of G are (2x1, 0) and of g are (0, 2y1)

So that PG = (2x1−x1)2+y12\sqrt{(2x_{1} - x_{1})^{2} + y_{1}^{2}}= x12+y12\sqrt{x_{1}^{2} + y_{1}^{2}} = PC

Pg = x12+(2y1−y1)2\sqrt{x_{1}^{2} + (2y_{1} - y_{1})^{2}}=x12+y12\sqrt{x_{1}^{2} + y_{1}^{2}}= PC and

Gg = (2x1)2+(2y1)2\sqrt{(2x_{1})^{2} + (2y_{1})^{2}}= 2x12+y12\sqrt{x_{1}^{2} + y_{1}^{2}}= 2PC.