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Q.1. The point of intersection of the tangents at the ends of the latus rectum of the parabola y^{2} = 4x is....... (1994  2 Marks)
Ans. ( –1, 0)
Sol. Given parabola is y^{2} = 4x; a = 1
Extremities of latus rectum are (1, 2) and (1, – 2) tangent to y^{2} = 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 x^{2} + y^{2} =1 and the hyperbola x^{2} – y^{2} =1. The equation of the ellipse, in the standard form, is............ (1996  2 Marks)
Ans.
Sol. Rough graph of x^{2} + y^{2} = 1 (circle) ...(1)
and x^{2} – y^{2} = 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.
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