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1. The line 2x + y = 1 is tangent to the hyperbola
If this line passes through the point of intersection of the nearest directrix and the xaxis, then the eccentricity of the hyperbola is (2010)
Ans. (2)
Sol. Intersection point of nearest directrix x = and xaxis
is
As 2x +y=1 passes through
Also y = –2x+1 is a tangent to
⇒ 4a^{2}– a^{2}e^{2} – 1=1 ⇒
⇒ 4e^{2 }–e^{4} +e^{2 }= 4 ⇒ e^{4} – 5e^{4} + 4=0
⇒ e^{2} = 4 ase>1 for hyperbola. ⇒ e= 2
2. Consider the parabola y^{2} = 8x . Let Δ_{1} be the area of the triangle formed by the end points of its latus rectum and the pointon the parabola and Δ_{2} be the area of the triangle formed by drawing tangents at P and at the end points of the latus rectum. Then is (2011)
Ans. (2)
Sol. Δ_{1} = Area of ΔPLL'
Equation of AB, y = 2x + 1 Equation of AC, y = x + 2 Equation of BC, – y = x + 2 Solving above equations we get A (1, 3), B (–1, –1), C (–2, 0)
3. Let S be the focus of the parabola y^{2} = 8x and let PQ be the common chord of the circle x^{2} + y^{2} – 2x – 4y = 0 and the given parabola. The area of the triangle PQS is (2012)
Ans. (4)
Sol. We observe both parabola y^{2} = 8x and circle x^{2} + y^{2} – 2x – 4y = 0 pass through origin
∴ One end of common chord PQ is origin. Say P(0, 0)
Let Q be the point (2t^{2}, 4t), then it will satisfy the equation of circle.
∴ 4t^{4} + 16t^{2} – 4t^{2} – 16t = 0 ⇒ t4 + 3t^{2} – 4t = 0 ⇒ t (t^{3 }+ 3t – 4) = 0
⇒ t (t – 1)(t^{2} + t – 4) = 0 ⇒ t = 0 or 1
For t = 0, we get point P, therefore t = 1 gives point Q as (2, 4).
We also observe here that P(0, 0) and Q(2, 4) are end points of diameter of the given circle and focus of the parabola is the point S(2, 0).
∴ Area of ΔPQS sq. units
4. A vertical line passing through the point (h, 0) intersects the
ellipse
at the points P and Q. Let the tangents to the ellipse at P and Q meet at the point R. If Δ(h) = area of the triangle then
(JEE Adv. 2013)
(a) g(x) is continuous but not differentiable at a
(b) g(x) is differentiable on R
(c) g(x) is continuous but not differentiable at b
(d) g(x) is continuous and differentiable at either (a) or (b) but not both
Ans. (9)
Sol. Vertical line x = h, meets the ellipse at
and
By symmetry, tangents at P and Q will meet each other at xaxis.
Tangent at P is
which meets xaxis at
Area of ΔPQR =
i.e.,
∴ Δ(h) is a decreasing function.
= 45 – 36 = 9
5. If the normals of the parabola y2 = 4x drawn at the end points of its latus rectum are tangents to the circle (x – 3)^{2} + (y + 2)^{2} = r^{2}, then the value of r^{2 }is (JEE Adv. 2015)
Ans. (2)
Sol. End points of latus rectum of y^{2} = 4x are (1, +2)
Equation of normal to y^{2 }= 4x at (1, 2) is y – 2 = –1(x – 1)
or x + y –3 = 0
As it is tangent to circle (x – 3)^{2} + (y + 2)^{2} = r^{2}
6. Let the curve C be the mirror image of the parabola y^{2} = 4x with respect to the line x + y + 4 = 0. If A and B are the points of intersection of C with the line y = –5, then the distance between A and B is (JEE Adv. 2015)
Ans. (4)
Sol. Let (t^{2}, 2t) be any point on y^{2} = 4x. Let (h, k) be the image of (t^{2}, 2t) in the line x + y + 4 = 0. Then
⇒ h = –(2t + 4) and k = –(t^{2} + 4)
For its intersection with, y = –5, we have –(t^{2} + 4) = –5 ⇒ t = +1
∴ A(–6, –5) and B(–2, –5) ∴ AB = 4.
7. Suppose that the foci of the ellipse
= 1 are (f_{1}, 0)and (f_{2}, 0) where f_{1} > 0 and f_{2} < 0. Let P_{1} and P_{2} be two parabolas with a common vertex at (0, 0) and with foci at (f_{1}, 0) and (2f_{2}, 0), respectively. Let T_{1} be a tangent to P_{1} which passes through (2f_{2}, 0) and T_{2 }be a tangent to P_{2} which passes through (f_{1}, 0). If m1 is the slope of T_{1} and m_{2} is the slope of T_{2}, then the value of
is (JEE Adv. 2015)
Ans. (4)
Sol. Ellipse :
⇒ a = 3, b =
∴ f_{1} = 2 and f_{2 }= –2 P_{1} : y^{2} = 8x and P_{2} : y^{2} = –16x
T_{1} : y = m_{1}x +
It passes through (–4, 0),
It passes through (2, 0)
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