JEE Advanced (Fill in the Blanks): Conic Sections | Chapter-wise Tests for JEE Main & Advanced PDF Download

Q.1. The point of intersection of the tangents at the ends of the latus rectum of the parabola y² = 4x is....... (1994 -  2 Marks)

Ans. ( –1,  0) 
Sol.  Given parabola is y² = 4x; a = 1
Extremities of latus rectum are (1, 2) and (1, – 2) tangent to y² = 4x at (1, 2) is y.² = 2 (x + 1) i.e.  y = x + 1 ...(1)
Similarly tangent at (1, – 2) is, y = – x – 1 ...(2)
Intersection pt. of these tangents can be obtained by solving (1) and (2), which is (– 1, 0).

Q.2. An ellipse has eccentricity and one focus at the point    Its one directrix is the common tangent, nearer to the point P, to the circle x² + y² =1 and the hyperbola x² – y² =1. The equation of the ellipse, in the standard form, is............ (1996 - 2 Marks)

Ans. 

Sol.  Rough graph  of x² + y² = 1 (circle) ...(1)
and x² – y² = 1 (hyperbola) ...(2)
is as shown below.

It is clear from graph that there are two common tangents to the curves (1) and (2) namely x = 1 and x = – 1 out of which x = 1 is nearer to pt. P.
Hence directrix of required ellipse is x – 1 = 0 Also e = 1/2, focus (1/2, 1) then equation of ellipse is given by

which is the standard equation of the ellipse.