Test Index

Hyperbola

Section: Mathematics

© examsnet.com

Question : 55 of 98

Marks: +1, -0

Let a line

L_1

be tangent to the hyperbola

\frac{x^2}{16} - \frac{y^2}{4} = 1

L_2

be the line passing through the origin and perpendicular to

L_1

. If the locus of the point of intersection of

L_1

L_2

(x^2+y^2)^2 = \alpha x^2 + \beta y^2

\alpha + \beta

is equal to____

[26-Jun-2022-Shift-2]

Your Answer:

Solution: 👈: Video Solution

Equation of

L_1

is

\frac{x \sec \theta}{4} - \frac{y \tan \theta}{2} = 1 \dots \text{ (i) }

Equation of line

L_2

is

\frac{x \tan \theta}{2} + \frac{y \sec \theta}{4} = 0 \dots \text{ (ii) }

\because

Required point of intersection of

L_1

and

L_2

is

(x_1, y_1)

then

\frac{x_1 \sec \theta}{4} - \frac{y_1 \tan \theta}{2} - 1 = 0 \dots \text{ (iii) }

and

\frac{y_1 \sec \theta}{4} - \frac{x_1 \tan \theta}{2} = 0

..... (iv)

From equations (iii) and (iv)

\sec \theta = \frac{4 x_1}{x_1^2 + y_1^2}

and

\tan \theta = \frac{-2 y_1}{x_1^2 + y_1^2}

\therefore

Required locus of

(x_1, y_1)

is

(x^2+y^2)^2 = 16 x^2 - 4 y^2

\therefore \alpha = 16, \beta = -4

\therefore \alpha + \beta = 12

© examsnet.com

Go to Question:

Prev Question

Next Question