Question

Which of the following statements are correct for oblique hyperbola $xy=8$
1. Equation of tangent at P(4, 2) is $x+2y=8$

2. Equation of normal at P(t) is $x{t}^3-yt=2\sqrt{2}(t4-1)$

• A. only 1
• B. only 2
• C. Both 1 & 2
• D. None of these

Answer 1

The correct option is C

Both 1 & 2

Equation of hyperbola is $S\equiv xy-8=0$

$S_1=4\times 2-8=0$

Since ,$S=0$ given point lies on the hyperbola.

Equation of the tangent is $T=0$

Given,$xy=8$

Multiplying 2 on both sides

$2xy=16$

$xy+xy=16$

Equation of tangent is

$x{y}_1+y{x}_1=16$

$X\times 2+y\times 4=16$

$x+2y=8$

Statement 1 is correct

point $p\left(t\right)$ is the parametric point on the parabola $xy=c^2$ which $ct,\frac{c}{t})$

For oblique hyperbola $xy=8$

parabola point is $(2\sqrt{2}t,\frac{2\sqrt{2}}{t})$

Equation ot tangent at $p\left(t\right)$

$\frac{x}{x_1}+\frac{y}{y_1}=2\frac{x}{2\sqrt{2t}}+\frac{y}{\frac{2\sqrt{2}}{t}}=2y\frac{t}{2\sqrt{2}}=-\frac{x}{2\sqrt{2t}}+2ty=-\frac{x}{t}+4\sqrt{2}y=\frac{-x}{t^2}+\frac{4\sqrt{2}}{t}\text{Slope of tangent}=\frac{-1}{t^2}\text{Slope of normal}=\frac{1}{\text{Slope of tangent}}=\frac{-1}{\frac{-1}{t^2}}=t^2$

Equation of normal with slope $t^2$ passes through the point p $(2\sqrt{2t},\frac{2\sqrt{2}}{t})$
$isy-\frac{2\sqrt{2}}{t}=t^2(x-2\sqrt{2t})ty-2\sqrt{2}=t^{3}x-t^4\times 2\sqrt{2}x{t}^3-ty=t^{4}2\sqrt{2}-2\sqrt{2}x{t}^3-ty=2\sqrt{2}(t^4-1)$

Statement 2 is also correct.