Question

Let C be the centre of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. The tangent at any point P on this hyperbola meets the straight lines $bx-ay=0$ and $bx+ay=0$ in the points Q and R respectively. Then CQ.CR is equal to:

• A. $a^2+b^2$
• B. $|a^2-b^2|$
• C. $\frac{1}{a^2}+\frac{1}{b^2}$
• D. $\frac{a^2{b}^2}{a^2+b^2}$

Answer 1

The correct option is A $a^2+b^2$

Given hyperbola : $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

equation of tangent at $P(x_1,y_1)$ is

T : $\frac{x{x}_1}{a^2}-\frac{y{y}_1}{b^2}=1$ ....... $\left(1\right)$

$(where\frac{{x_1}^2}{a^2}-\frac{{y_1}^2}{b^2}=1)$ ......... $\left(2\right)$

Point of intersection of tangent T and lines

of $y=\frac{b}{a}x$ $\Rightarrow \frac{x{x}_1}{a^2}-\frac{y_1}{ab}=1$

$\Rightarrow x=(a^2+\frac{y_{1}a}{b})\frac{1}{x_1}$

$\&y=(ab+y_1)\frac{1}{x_1}$

Q : $((a^2+\frac{y_{1}a}{b})\frac{1}{x_1},(ab+y_1\left)\frac{1}{x_1}\right)$

of $y$=$\frac{-b}{a}x$ $\Rightarrow x=(a^2-\frac{y_{1}a}{b})\frac{1}{x_1}\Rightarrow y=(ab-y_1)\frac{1}{x_1}$

R : $((a^2-\frac{y_{1}a}{b})\frac{1}{x_1},(ab-y_1\left)\frac{1}{x_1}\right)$

To find $\overline{CQ}.\overline{CR}=\sqrt{{\left[(a^2+\frac{y_{1}a}{b})\frac{1}{x_1}\right]}^2+{\left[\right(ab+y_1\left)\frac{1}{x_1}\right]}^2}.\sqrt{{\left[(a^2-\frac{y_{1}a}{b})\frac{1}{x_1}\right]}^2+{\left[\right(ab-y_1\left)\frac{1}{x_1}\right]}^2}$

$=\sqrt{\frac{1}{{x_1}^2}[a^4-\frac{{y_1}^2{a}^2}{b^2}]+\frac{1}{{x_1}^2}{[a^2{b}^2-{y_1}^2]}^2+\frac{1}{{x_1}^4}\left[2(a^{3}b-{y_1}^2\frac{a}{b})\right]}$

$=\sqrt{\frac{1}{{x_1}^4}[{x_1}^2(a^4-{y_1}^2\frac{a^2}{b^2}-{y_1}^2+a^2{b}^2)+2(a^{3}b-{y_1}^2\frac{a}{b})]}$

$\Rightarrow$ Now put $(x_1,y_1)as(a\sec \alpha ,b\tan \alpha )$ and they do give

$\Rightarrow \sqrt{a^4+2a^2{b}^2+b^2}$

$=a^2{+b}^2$