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The n umber of values of c such that the straight line y = 4x + c touches the curve (x^{2}/4) + y^{2} = 1 is (1998  2 Marks)
The given curve is (an ellipse) and given line is y = 4x + c.
We know that y = mx + c touches the ellipse
Here
∴ two values of c exist
If P = (x, y), F_{1} = (3, 0), F_{2} = (–3, 0) and 16x^{2} + 25y^{2 }= 400, then PF_{1} +PF_{2 }equals (1998  2 Marks)
The ellipse can be written as,
Here a^{2} = 25, b^{2} = 16, but b^{2} = a^{2} (1 – e2)
⇒ 16/25 = 1– e^{2} ⇒ e^{2} =1 – 16/25 = 9/25 ⇒ e = 3/5
Foci of the ellipse are (+ ae, 0) = (± 3, 0), i.e., F_{1} and F_{2}
∴ We have PF_{1} + PF_{2} = 2a = 10 for every point P on the ellipse.
On the ellipse 4 x^{2} + 9y^{2 }=1 , the points at which the tangents are parallel to the line 8x = 9y are(1999  3 Marks)
Let be the tangent to
where
and pts of contact are
or
The equations of th e common tangents to the parabola y = x^{2} and y = – (x – 2)^{2 }is/are (2006  5M, –1)
if y = mx + c is tangent to y = x^{2} then x^{2} – mx – c = 0 has equal roots
⇒ m^{2} + 4c = 0 ⇒ c
is tangent to y = x^{2 }
∴ This is also tangent to y = – (x – 2)^{2}
= x^{2} + 4x – 4
⇒ x^{2} + (m – 4)x += 0
has equal roots⇒ m^{2} – 8m + 16 = – m^{2} + 16
⇒ m = 0, 4 ⇒ y = 0 or y = 4x – 4 are the tangents.
Let a hyperbola passes through the focus of the ellipse . The transverse and conjugate axes of this hyperbola coincide with the major and minor axes of the given ellipse, also the product of eccentricities of given ellipse and hyperbola is 1, then (2006  5M, –1)
For the given ellipse
⇒ Eccentricity of hyperbola
Let the hyperbola be = 1 then
= 1
As it passes through focus of ellipse
i.e. (3, 0)
∴ we get A^{2} = 9 ⇒ B^{2} = 16
∴ Equation of hyper bola isfocus of hyperbola is (5, 0), vertex of hyperbola is (3, 0).
Let P(x_{1}, y_{1}) and Q(x_{2}, y_{2}), y_{1} < 0, y_{2} < 0, be the end points of the latus rectum of the ellipse x^{2 }+ 4y^{2} = 4. The equations of parabolas with latus rectum PQ are (2008)
Given ellipse is x^{2} + 4y^{2} = 4
a = 2, b = 1
As per question
∴ PQ = 2
Now if PQ is the length of latus rectum to be found, then PQ = 4a =
Also as PQ is horizontal, parabola with PQ as latus rectum can be upward parabola (with vertex at A) or down ward parabola (with vertex at A').
For upward parabola,
AR = a = ∴ Coordinates of A =
So equation of upward parabola is given by
or
For downward parabola A'R = a =
∴ Coordinates of A' =
So equation of downward parabola is given by
or
∴ the equation of required parabola is given by equation (1) or (2).
In a triangle ABC with fixed base BC, the vertex A moves such that
cosB + cosC = .
If a, b and c denote the lengths of the sides of the triangle opposite to the angles A, B and C, respectively, then
In ΔABC , given that
But in a triangle sin
Applying componendo and dividendo, we get
⇒ a + b + c = 3a or b + c = 2a
i.e. AC+ AB = constant
(∵ Base BC = a is given to be constant) ⇒ A moves on an ellipse.
The tangent PT and the normal PN to the parabola y^{2} = 4ax at a point P on it meet its axis at points T and N, respectively.The locus of the centroid of the triangle PTN is a parabola whose(2009)
Let P(at^{2} , 2at) be any point on the parabola y^{2} = 4ax.
Then tangent to parabola at P is
which meets the axis of parabola i.e xaxis at T (– at^{2}, 0).
Also normal to parabola at P is tx + y = 2at+ at^{3} which meets the axis of parabola at N (2a + at^{2} , 0)
Let G (x, y) be the centriod of Δ PTN , then
Eliminating t from above, we get the locus of centriod G as
which is a parabola with vertex directrix as
latus r ectum as and
focus as (a, 0).
An ellipse intersects the hyperbola 2x^{2} – 2y^{2} = 1 orthogonally.The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinate axes, then (2009)
The given hyperbola is
...(1)
which is a rectangular hyperbola (i.e. a = b)
Let the ellipse be
Its eccentricity =
So, the equation of ellipse becomes x^{2} + 2 y^{2 }= a^{2 }...(2)
Let the hyperbola (1) and ellipse (2) intersect each other at P( x_{1} ,y_{1}).
Then slope of hyperbola (1) at P is given by
and that of ellipse (2) at P is
As the two curves intersect orthogonally,
∴ m_{1}m_{2} =1
Also P ( x_{1},y_{1}) lies on x^{2} y^{2 }=
...(ii)
Solving (i) and (ii), we get
Also P ( x_{1},y_{1}) lies on ellipse x^{2} + 2y^{2 }= a^{2}
1 + 1 = a^{2} or a^{2} =2
∴ The required ellipse is x^{2} + 2y^{2 }= 2 whose foci
Let A and B be two distinct points on the parabola y^{2} = 4x. If the axis of the parabola touches a circle of radius r having AB as its diameter, then the slope of the line joining A and B can be
(2010)
Given parabola y^{2} = 4x
Let A(t_{1}^{2} ,2t_{1} ) and B(t_{2}^{2} ,2t_{2}) Then centre of circle drawn with AB as diameter is
As circle touches xaxis
∴ r =  t_{1} + t_{2} ⇒ t_{1} + t_{2} = ±r
Also slope of AB
Let the eccentricity of the hyperbol a abbereciprocal to that of the ellipse x^{2 }+ 4y^{2} = 4. If the hyperbola passes through a focus of the ellipse, then (2011)
For x^{2} + 4y^{2} = 4 or
As per question,
focus of ellipse is (±, 0) As hyperbola passes through (±, 0)
∴ b = 1 and focus of hyperbola (± 2, 0)
∴ Equation of hyperbola is
or x^{2} – 3y^{2} = 3
Let L be a normal to the parabola y^{2} = 4x. If L passes through the point (9, 6), then L is given by (2011)
The equation of normal to y^{2} = 4x is y = mx – 2m – m^{3} As it passes through (9, 6)
∴ 6 = 9m – 2m – m^{3} ⇒ m^{3} – 7m + 6 = 0 ⇒ (m – 1) (m^{2} + m – 6) = 0
⇒ (m – 1) (m + 3) (m – 2) = 0 ⇒ m = 1, 2, –3
∴ Normal is y = x – 3 or y = 2x – 12 or y = – 3x + 33
∴ a, b, d are the correct option.
Tangents are drawn to the hyperbola parallel to the straight line 2x – y = 1. The points of contact of the tangents on the hyperbola are (2012)
KEY CONCEPT : If slope of tangent is m, then equations of tangents to hyperbola
= are
with the points of contact
∵ Tangent to hyperbola = is parallel to 2x – y = 1, therefore slope of tangent = 2
∴ Points of contact are
i.e.
Let P and Q be distinct points on the parabola y^{2} = 2x such that a circle with PQ as diameter passes through the vertex O of the parabola. If P lies in the first quadrant and the area of the triangle ΔOPQ is 3, then which of the following is (are) the coordinates of P? (JEE Adv. 2015)
Let point P in first quadrant, lying on parabola y^{2} = 2x
be Let Q be the point Clearly a > 0.
∵ PQ is the diameter of circle through P, O, Q
∴ ∠POQ = 90° = –1 ⇒ ab = –4
⇒ b is negative.
Also ar. ΔPOQ = 3
⇒
(using ab = –4)
As a is positive and b is negative, we have a – b = 3
(using ab = –4)
⇒ a^{2} – 3a + 4 = 0 ⇒ a^{2} – 2a – a + 4 = 0
⇒ (a – 2) (a – ) = 0 ⇒ a = 2 ,
∴ Point P can be or
i.e. (4, 2) or (1, )
Let E_{1 }and E_{2} be two ellipses whose centers are at the origin.
The major axes of E_{1} and E_{2 }lie along the xaxis and the yaxis, respectively. Let S be the circle x^{2} + (y – 1)^{2} = 2. The straight line x + y = 3 touches the curves S, E_{1} and E_{2} at P, Q and R respectively. Suppose that PQ = PR =
If e_{1} and e_{2} are the eccentricities of E_{1 }and E_{2}, respectively, then the correct expression(s) is (are) (JEE Adv. 2015)
Let E_{1} := 1 where a > b
and E_{2} := 1 where c < d
Also S : x^{2} + (y – 1)^{2} = 2
Tangent at P(x1, y1) to S is x + y = 3
To find point of contact put x = 3 – y in S.
We get P(1, 2) Writing eqn of tangent in parametric form
and
and
eqn of tangent to E_{1} at Q is
which is identical to
⇒ a^{2} = 5 and b^{2} = 4 ⇒
eqn of tangent to E^{2} at R is
identical to
⇒ c^{2} = 1, d^{2} = 8 ⇒
Consider the hyperbola H : x^{2} – y^{2} = 1 and a circle S with center N(x^{2}, 0). Suppose that H and S touch each other at a point P(x_{1}, y_{1}) with x_{1} > 1 and y_{1} > 0. The common tangent to H and S at P intersects the xaxis at point M. If (l, m) is the centroid of the triangle PMN, then the correct expression(s) is(are) (JEE Adv. 2015)
H : x^{2} – y^{2 }= 1 S : Circle with centre N(x_{2}, 0) Common tangent to H and S at P(x_{1}, y_{1}) is
xx_{1} – yy_{1} = 1 ⇒
Also radius of circle S with centre N(x_{2}, 0) through point of contact (x_{1}, y_{1}) is perpendicular to tangent
⇒ x_{1} = x_{2} – x_{1} or x_{2 }= 2x_{1}
M is the point of intersection of tangent at P and xaxis
∵ Centroid of ΔPMN is (ℓ, m)
and y_{1} = 3m
Using x_{2} = 2x_{1},
Also (x_{1}, y_{1}) lies on H,
The circle C_{1} : x2 + y2 = 3, with centre at O, intersects the parabola x^{2} = 2y at the point P in the first quadrant. Let the tangent to the circle C_{1}, at P touches other two circles C_{2} and C_{3} at R_{2 }and R_{3}, respectively. Suppose C_{2} and C_{3} have equal radii 2 and centres Q_{2} and Q_{3}, respectively. If Q_{2} and Q_{3} lie on the y–axis, then (JEE Adv. 2016)
C_{1} : x^{2} + y^{2} = 3 ..(i)
parabola : x^{2 } = 2y ...(ii)
Intersection point of (i) and (ii) in first quadrant y^{2} + 2y – 3 = 0 ⇒ y = 1 (∵y ≠ 3)
∴x =
P( ,1)
Equation of tangent to circle C_{1} at P is x + y3 = 0
Let centre of circle C_{2} be (0, k); r = 2
k = 9 or –3
∴ Q_{2} (0, 9), Q_{3} (0, –3)
(a)Q_{2} Q_{3} = 12
(b) R_{2}R_{3} = length of transverse common tangent
(c) Area × length of ⊥ from originto tangent
(d) Area distance of P from
y–axis =
Let P be the point on the parabola y^{2} = 4x wh ich is at the shortest distance from the center S of the circle x^{2} + y^{2 }– 4x –16y + 64 = 0. Let Q be the point on the circle dividing the line segment SP internally. Then (JEE Adv. 2016)
Let point P on parabola y^{2} = 4x be (t^{2}, 2t)
∵ PS is shortest distance, therefore PS should be the normal to parabola.
Equation of normal to y^{2} = 4x at P (t^{2}, 2t) is y – 2t = – t(x – t^{2})
It passes through S(2, 8)
∴ 8 – 2t = – t(2 – t^{2}) ⇒ t^{3} = 8 or t = 2
∴ P(4, 4)
Also slope of tangent to circle at
Equation of normal at t = 2 is 2x + y = 12
Clearly xintercept = 6
SP = and SQ = r = 2
∴ Q divides SP in the ratio SP : PQ
Hence a, c , d are the correct options.
132 docs70 tests

132 docs70 tests
