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Two common tangents to the circle x^{2} + y^{2 }= 2a^{2} and parabola y^{2} = 8ax are [2002]
Any tangent to the parabola y^{2} = 8ax is...(i)
If (i) is a tangent to the circle, x^{2} + y^{2} = 2a^{2} then,
⇒ m^{2}(1 + m^{2}) = 2 ⇒ (m^{2} + 2)(m^{2} – 1) = 0 ⇒ m = ± 1.
So from (i), y = ± (x + 2a).
The normal at the point (bt_{1}^{2} , 2bt^{1}) on a parabola meets the parabola again in the point (bt_{2}^{2}, 2bt_{2}) , then [2003]
Equation of the normal to a parabola y^{2} = 4bx at point
(bt_{1}^{2},2bt_{1}) is y = – t_{1} x + 2bt_{1}+ bt_{1}^{3}
As given, it also passes through (bt_{2}^{2} ,2bt_{2}) then
⇒ t_{2 }+ t_{1 }=
The foci of the ellipse and the hyper bola coincide. Then the value of b^{2} is [2003]
∴ Foci = ( ±3 , 0)
∴ foci of ellipse = foci of hyperbola
∴ for ellipse ae = 3 but a = 4,
∴
Then b^{2} = a^{2} (1e2) ⇒ b^{2 }=
If a ≠ 0 an d the lin e 2bx + 3cy + 4d=0 passes through the points of intersection of the parabolas y^{2 }= 4ax and x^{2 }= 4ay, th en [2004]
Solving equations of parabolas y^{2} = 4ax and x^{2 }= 4ay
we get (0, 0) and ( 4a, 4a)
Substituting in the given equation of line 2bx + 3cy + 4d = 0,
we get d = 0 and 2b +3c = 0
⇒ d^{2} + (2b + 3c)^{2 }= 0
The eccentricity of an ellipse, with its centre at the origin, is. If one of the directrices is x = 4 , then the equation ofthe ellipse is: [2004]
e = Directrix ,
Equation of elhipe is 3x^{2} + 4y^{2} = 12
Let P be the point ( 1, 0 ) and Q a point on the locus y^{2} = 8x .The locus of mid point of PQ is [2005]
P = (1, 0) Q = (h, k)
Such that K^{2} = 8h Let (α, β) be the midpoint of PQ
2α  1 = h 2β = k.
(2β) ^{2} = 8(2α  1)
⇒ β^{2 }= 4α  2
⇒ y^{2} 4x+2=0.
The locus of a point P (α, β) moving under the condition th at the line y = αx + β is a tangen t to the hyper bola is [2005]
Tangent to the hyperbola is
Given that y = αx + β is the tangent of hyperbola
⇒ m = α and a^{2} m^{2} b^{2} =β^{2}
∴ a^{2}α^{2} b^{2} =β^{2}
Locus is a^{2} x^{2}  y^{2 }= b^{2} which is hyperbola.
An ellipse has OB as semi minor axis, F and F ' its focii and the angle FBF ' is a right angle. Then the eccentricity of the ellipse is [2005]
⇒ 2(a^{2}e^{2} + b^{2})=4a^{2}e^{2} ⇒
Also e^{2} = 1 b^{2} / a^{2} =1e^{2}
The locus of the vertices of the family of parabol as [2006]
Given parabola is
∴ Vertex of parabola is
To find locus of this vertex,
64xy = 105
which is the required locus.
In an ellipse, the distance between its foci is 6 and minor axis is 8. Then its eccentricity is [2006]
2ae =6 ⇒ ae = 3 ; 2b =8 ⇒ b = 4
b^{2} = a^{2}(1e^{2}) ;16 = a^{2} a^{2}e^{2} ⇒ a^{2} = 16 + 9= 25
⇒ a = 5
Angle between the tangents to the curve y = x^{2 } 5x+6 at the points (2, 0) and (3, 0) is [2006]
∴ m_{1} = (2x  5)(2, 0) =1 ,
m_{2} = (2x  5)_{(3, 0)}=1 ⇒ m_{1}m_{2} =1
i.e. the tangents are perpendicular to each other.
For the Hyper bola which of the following remains constant when a varies = ? [2007]
Given, equation of hyperbola is
We know that the equation of hyperbola is
Here, a^{2} = cos^{2}α and b^{2} = sin^{2}α
We know that, b^{2} = a^{2} (e^{2 } 1)
⇒ sin^{2}α = cos^{2}α(e^{2 } 1)
⇒ sin^{2}α + cos^{2}α = cos^{2}α.e^{2}
⇒ e^{2} = 1 + tan^{2}α = sec^{2}α ⇒ e = secα
Coordinates of foci are (± ae, 0) i.e. ( ± 1, 0)
Hence, abscissae of foci remain constant when α varies.
The equation of a tangent to the parabola y^{2} = 8x is y = x + 2. The point on this line from which the other tangent to the parabola is perpendicular to the given tangent is [2007]
Parabola y^{2} = 8x Y
We know that the locus of point of intersection of two perpendicular tangents to a parabola is its directrix.
Point must be on the directrix of parabola
∵ equation of directrix x + 2 = 0 ⇒ x = –2
Hence the point is (–2, 0)
The normal to a curve at P(x, y) meets the xaxis at G. If the distance of G from the origin is twice the abscissa of P, then the curve is a [2007]
Equation of normal at P(x, y) is Y – y =
Coordinate of G at X axis is (X, 0) (let)
⇒
∴ Coordinate of G
Given distance of G from origin = twice of the abscissa of P.
∵ distance cannot be –ve, therefore abscissa x should be +ve
On Integrating ⇒ ⇒ x^{2} – y^{2} = –2c_{1}
∴ the curve is a hyperbola
A focus of an ellipse is at the origin. The directrix is the line x = 4 and the eccentricity is . Then the length of the semimajor axis is [2008]
Perpendicular distance of directrix from focus
∴ Semi major axis = 8/3
A parabola has the origin as its focus and the line x = 2 as the directrix. Then the vertex of the parabola is at [2008]
Vertex of a parabola is the mid point of focus and the point
where directrix meets the axis of the parabola.
Here focus is O(0, 0) and directxix meets the axis at B(2, 0)
∴ Vertex of the parabola is (1, 0)
The ellipse x^{2} + 4y^{2 }= 4 is inscribed in a rectangle aligned with the coordinate axes, which in turn is inscribed in another ellipse that passes through the point (4, 0). Then the equation of the ellipse is : [2009]
The given ellipse is
So A = (2, 0) and B= (0, 1) If PQRS is the rectangle in which it is inscribed, then P = (2, 1).
be the ellipse circumscribing the rectangle PQRS.
Then it passes through P (2,1 )
Also, given that, it passes through (4, 0)
⇒ b^{2} = 4/3 [substituting a^{2} = 16 in eq^{n }(a)]
∴ The required ellipse is or x^{2} + 12y^{2} =16
If two tangents drawn from a point P to the parabola y^{2} = 4x are at right angles, then the locus of P is [2010]
The locus of perpendicular tangents is directrix i.e., x =1
Equation of the ellipse wh ose axes ar e th e axes of coordinates and which passes through the point (–3, 1) and has eccentricity is [2011]
Let the ellipse be
It passes through (– 3, 1) so
Also, b^{2} =a^{2 }(1 2 / 5) ⇒ 5b^{2 }=3a^{2} ...(ii)
Solving (i) and (ii) we get a^{2 }=
So, the equation of the ellipse is 3x^{2} + 5y^{2 }= 32
Statement1 : An equation of a common tangent to the parabola y^{2} = 16x and the ellipse 2x^{2} + y^{2 }=4 is y = 2x+ 2
Statement2 : If the line is a common tan gent to the parabola y^{2} = 16x and the ellipse 2x^{2} + y^{2} = 4, then m satisfies m^{4} + 2m^{2} = 24 [2012] [2012]
Given equation of ellipse is 2x^{2} + y^{2} = 4
Equation of tangent to the ellipse
...(1)
(∵ equation of tangent to the ellipse
is y = mx + c where
Now, Equation of tangent to the parabola
...(2)
(∵ equation of tangent to the parabola y^{2} = 4ax is
On comparing (1) and (2), we get
Squaring on both the sides, we get 16 (3) = (2m^{2} + 4) m^{2 }
⇒ 48 = m^{2} (2m^{2} + 4) ⇒ 2m^{4} + 4m^{2} – 48 = 0 ⇒ m^{4 }+ 2m^{2} –24 = 0
⇒ (m^{2} + 6)(m^{2} – 4) = 0 ⇒ m^{2}= 4 (∵ m^{2} ≠ – 6) ⇒ m = ± 2
⇒ Equation of common tangents are
Thus, statement1 is true.
Statement2 is obviously true.
An ellipse is drawn by taking a diameter of the circle (x – 1)^{2} + y^{2} = 1 as its semiminor axis and a diameter of the circle x^{2} + (y – 2)^{2} = 4 is semimajor axis. If the centre of the ellipse is at the origin and its axes are the coordinate axes, then the equation of the ellipse is : [2012]
Equation of circle is (x – 1)^{2} + y^{2} =1 ⇒ radius = 1 and diameter = 2
∴ Length of semiminor axis is 2.
Equation of circle is x^{2} + (y – 2)^{2} = 4 = (2)^{2} ⇒ radius = 2 and diameter = 4
∴ Length of semi major axis is 4 We know, equation of ellipse is given by
The equation of the circle passing through the foci of the ellipse and having centre at (0, 3) is [JEE M 2013]
From the given equation of ellipse, we have
a = 4, b = 3,
Now, radius of this circle =
⇒ Focii = (±, 0)
Now equation of circle is (x – 0)^{2} + (y – 3)^{2} = 16
x^{2} + y^{2} – 6y – 7 = 0
Given : A circle, 2x^{2} + 2y^{2} = 5 and a parabola,
Statement1 : An equation of a common tangent to these curves is
Statement2 : If the line, is theircommon tangent, then m satisfies m^{4} – 3m^{2} + 2 = 0. [JEE M 2013]
Let common tangent be
Since, perpendicular distance from centre of the circle to the common tangent is equal to radius of the circle,
therefore
On squaring both the side, we get m^{2} (1 + m^{2}) = 2
⇒ m^{4} + m^{2 }– 2 = 0
⇒ (m^{2} + 2)(m^{2} – 1) = 0
⇒ m = ± 1
both statements are correct as m = ±1 satisfies the given equation of statement2.
The locus of the foot of perpendicular drawn from the centre of the ellipse x^{2} + 3y^{2 }= 6 on any tangent to it is [JEE M 2014]
Given eqn of ellipse can be written as
a^{2} = 6, b^{2 }= 2
Now, equation of any variable tangent is
where m is slope of the tangent So, equation of perpendicular line drawn from centre to tangent is
...(ii)
Eliminating m, we get
(x^{4} + y^{4} + 2x^{2} y^{2}) = a^{2}x^{2}+ b^{2}y^{2}
⇒ ( x^{2} + y^{2})^{2 }= a^{2} x^{2 }+ b^{2}y^{2}
The slope of the line touching both the parabolas y^{2} = 4x and x^{2} =32y is [JEE M 2014]
Let tangent to y^{2} = 4x be y =
Since this is also tangent to x^{2} = – 32y
Now, D = 0
Let O be the vertex and Q be any point on the parabola, x^{2} = 8y. If the point P divides the line segment OQ internally in the ratio 1 : 3, then locus of P is : [JEE M 2015]
Let P(h, k) divides OQ in the ratio 1 : 3
Let any point Q on x^{2} = 8y is (4t, 2t^{2}).
Then by section formula
⇒ 2k =h^{2}
Required locus of P is x^{2} = 2y
The normal to the curve, x^{2} + 2xy – 3y^{2} = 0, at (1, 1) [JEE M 2015]
Given curve is x^{2} + 2xy – 3y^{2} = 0 ...(1)
Differentiate w.r.t. x,
Equation of normal at (1, 1) is y = 2 – x ...(2)
Solving eq. (1) and (2), we get x = 1 , 3
Point of intersection (1, 1), (3, –1)
Normal cuts the curve again in 4th quadrant.
The area (in sq. units) of the quadrilateral formed by the tangents at the end points of the latera recta to the ellipse is : [JEE M 2015]
The end point of latus rectum of ellipse
in first quadrant is and the tangent at this point intersects xaxis at and yaxis at (0, a).
The given ellipse is
Then a^{2} = 9, b^{2} = 5 ⇒
∴ end point of latus rectum in first quadrent is L (2, 5/3)
Equation of tangent at L is
It meets xaxis at A (9/2, 0) and yaxis at B (0, 3)
∴ Area of ΔOAB
By symmetry area of quadrilateral
= 4 × (Area ΔOAB) = sq. units.
Let P be the point on the parabola, y^{2} = 8x which is at a minimum distance from the centre C of the circle, x^{2} + (y + 6)^{2} = 1.Then the equation of the circle, passing through C and having its centre at P is: [JEE M 2016]
Minimum distance
⇒ perpendicular distance Eq^{n} of normal at p(2t^{2}, 4t) y = –tx + 4t + 2t^{3}
It passes through C(0, –6) ⇒ t^{3} + 2t + 3 = 0 ⇒ t = – 1
Centre of new circle = P(2t^{2} , 4t) = P(2,4)
Radius
∴ Equation of the circle is
⇒ x^{2} + y^{2} – 4x + 8y + 12 = 0
The eccentricity of the hyperbola whose length of the latus rectum is equal to 8 and the length of its conjugate axis is equal to half of the distance between its foci, is : [JEE M 2016]
⇒ 4b^{2} = a^{2}e^{2} ⇒ 4a^{2}(e^{2} 1)= a^{2}e^{2} ⇒ 3e^{2} = 4
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