Test Index

Hyperbola

Section: Mathematics

© examsnet.com

Question : 11 of 34

Marks: +1, -0

The tangents drawn to the hyperbola

5 x^2-9 y^2=90

through a variable point

P

make the angles

\alpha

\beta

with its transverse axis. If

\alpha, \beta

are the complementary angles, then the locus of P is

[AP EAPCET 21 May 2025 S2]

$x^2+y^2=8$
$x^2-y^2=8$
$x^2-y^2=28$
$x^2+y^2=28$

Solution:

Given, hyperbola

5 x^2-9 y^2=90

\;\frac{x^2}{18}-\;\frac{y^2}{10}=1

Let equation of tangent to the ellipse be

y=m x \pm \sqrt{18 m^2-10}

or

(y-m x)^2=18 m^2-10

\therefore \;y^2+m^2 x^2-2 m x y-18 m^2+10=0

\;m^2 (x^2-18) -2 m x y+y^2+10=0

Given tangent makes

\alpha

and

\beta

angles with transverse axis are complementary

\; \because \;\; \alpha+\beta=\;\frac{\pi}{2}

\;\tan (\alpha+\beta)=\;\text{ not defined }\;

\;\;\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}=\;\text{ not defined }\;

\;\;\text{ And }\; m_1=\tan \alpha \;\; m_2=\tan \beta

\;m_1+m_2=\;\frac{2 x y}{x^2-18} \;\text{ and }\; m_1 m_2=\;\frac{y^2+10}{x^2-18}

\; \because \;\; 1-m_1 m_2=0 \Rightarrow m_1 m_2=+1

\; \;\; \;\frac{y^2+10}{x^2-18}=+1

\;\Rightarrow \;\; y^2+10=+x^2-18

\;\Rightarrow \;\; y^2=x^2-28 \;\text{ or }\; x^2-y^2=28

© examsnet.com

Go to Question:

Prev Question

Next Question