Question

Solution

Correct option is

16x2 + 10xy + y2 = 2

P(h, k) be any point on the variable line with slope = 4. Hence it equation is y – k m(x – h)                                        …… (1)

m = 4, let this line (1) cut the hyperbola xy = 1            …… (2)

at A(x1y1) and B(x2y2). By assumption, P divides AB in the ratio 1 : 2 then Thus ⇒ 2x1 + x2 = 3h , 2y1 + y2 = 3k                             …... (3)

By (2), , put this in (1), we get ⇒ 4x2 – (4h – k) x – 1 = 0                                               …… (4)

x1x2 are roots of (4) …… (5) …… (6)

Solving (3) and (4), for x1 and x2, we get Putting these in (6) Locus of (h, k) = 16x2 + 10xy – 2 = 0.

SIMILAR QUESTIONS

Q1

Find the equations of tangents to the hyperbola x2 – 4y = 36 which are perpendicular to the line x – y + 4 = 0

Q2

Find the coordinates of foci, the eccentricity and latus rectum. Determine also the equation of its directrices for the hyperbola

4x2 – 9y2 =36.

Q3

Find the distance from A(4, 2) to the points in which the line 3x – 5= 2 meets the hyperbola xy = 24. Are these points on the same side of A?

Q4

The asymptotes of the hyperbola makes an angle 600 with x-axis. Write down the equation of determiner conjugate to the diameter y = 2x.

Q5

Two straight lines pass through the fixed points and have gradients whose product is k > 0. Show that the locus of the points of intersection of the lines is a hyperbola.

Q6

Find the equation of the triangles drawn from the point (–2, –1) to the hyperbola 2x2 – 3y2 = 6.

Q7

Find the range of ‘a’ for which two perpendicular tangents can be drawn to the hyperbola from any point outside the hyperbola .

Q8

Find the hyperbola whose asymptotes are 2x – y = 3 and 3x + y – 7 = 0 and which passes through the point (1, 1).

Q9

The locus of a variable point whose distance from (–2, 0) is times its distance from the line , is

Q10

Let P and , where , be two points on the hyperbola . If (h, k) is the point of intersection of the normal’s at P and Q, then k is equal to