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The equation represents(1981  2 Marks)
Given that
As r > 1
∴ 1 – r < 0 and 1 + r > 0
∴ Let 1 – r = – a^{2}, 1 + r = b^{2}, then we get
which is not possible for any real values of x and y.
Each of the four inequalties given below defines a region in the xy plane. One of these four regions does not have the following property. For any two points (x_{1}, y_{1}) and (x_{2}, y_{2}) in th e region , the point is also in the region. The inequality defining this region is (1981  2 Marks)
(a) x^{2} + 2y^{2} ≤ 1 represents interior region of an ellipse where on taking any two pts the mid pt of that segment will also lie inside that ellipse
(b) Max {  x ,  y  } ≤ 1 ⇒  x  ≤ 1,  y  ≤ 1 ⇒ – 1 ≤ x ≤ 1 and – 1 ≤ y ≤ 1
which represents the interior region of a square with its sides x = ± 1 and y = ± 1 in which for any two pts, their mid pt also lies inside the region.
(c) x^{2} – y^{2} ≥ 1 repr esents the exterior r egion of hyperbola in which if we take two points (2, 0) and (– 2, 0) then their mid pt (0, 0) does not lie in the same region (as shown in the figure.)
(d) y^{2} ≤ x represents interior region of parabola in which for any two pts, their mid point also lie inside the region.
The equation 2x^{2} + 3y^{2} – 8x – 18y + 35 = k represents (1994)
We have 2x^{2} + 3y^{2} – 8x – 18y + 35 = k
⇒ 2 (x – 2)^{2} + 3 (y – 3)^{2} = k
For k = 0, we get 2 (x – 2)^{2} + 3 (y – 3)^{2} = 0 which represents the point (2, 3).
Let E be the ellipse and C be the circle x^{2 }+ y^{2 }= 9. Let P and Q be the points (1, 2) and (2, 1) respectively. Then (1994)
Since 1^{2} + 2^{2} = 5 < 9 and 2^{2} + 1^{1} = 5 < 9 both P and Q lie inside C.
Also and P lies outside E and Q lies inside E. Thus P lies inside C but outside E.
Consider a circle with its centre lying on the focus of the parabola y^{2} = 2px such that it touches the directrix of the parabola. Then a point of intersection of the circle and parabola is (1995S)
The focus of parabola y^{2 }= 2px is and directrix
x = – p/2
In the figure, we have supposed that p > 0]
∴ Centre of circle is and radius
∴ Equation of circle is
For pts of intersection of y^{2} = 2px ...(i)
and 4x^{2} + 4y^{2} – 4px – 3p^{2} = 0 ...(ii)
can be obtained by solving (i) and (ii) as follows 4x^{2} + 8px – 4px – 3p^{2} = 0 ⇒ (2x + 3p) (2x – p) = 0
⇒ y^{2} = – 3p^{2} (not possible), p^{2} ⇒ y = + p
∴ Required pts are (p/2, p), (p/2, – p)
The radius of the circle passing through the foci of theellipse and having its centre at (0, 3) is (1995S)
For ellipse = 1, a = 4, b = 3
∴ Foci are
Centre of circle is at (0, 3) and it passes through
, therefore radius of circle =
Let P (a secθ, b tanθ) and Q (a secφ, b tanφ), where q + f = π/ 2, be two points on the hyperbolaIf (h, k) is the point of intersection of the normals at P and Q, then k is equal to (1999  2 Marks)
KEY CONCEPT : Equation of the normal to the hyperbola
at the point (a secα, b tanα) is given by ax cosα + by cotα = a^{2} + b^{2}
Normals at
where and these pass through (h, k)
∴ ah cosθ + bk cotθ = a^{2} + b^{2} ah sinθ + bk tanθ = a^{2 }+ b^{2}
Eliminating h, bk (cotθ sinθ – tan cosθ ) = (a^{2 }+ b^{2})
(sinθ – cosθ ) or k = – (a^{2} + b^{2})/b
If x = 9 is the chord of contact of the hyperbola x^{2} – y^{2} = 9, then the equation of the corresponding pair of tangents is (1999  2 Marks)
Chord x = 9 meets x^{2} – y^{2} = 9 at (9,6 ) and (9, –6)
at which tangents are 9x –6y =9 and 9x+6y=9
or 3x –2y – 3 =0 and 3x + 2y – 3=0
∴ Combined equation of tangents is (3x –2y –3)(3x + 2y – 3)=0
or9x^{2} – 8y^{2} – 18x + 9 = 0
Th e curve described par ametr ically by x = t^{2} + t + 1, y = t^{2} – t + 1 represents (1999  2 Marks)
KEY CONCEPT
The equation ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0 represents a parabola if Δ ≠ 0 and h^{2} = ab where Δ = abc + 2fgh – af^{ 2} – bg^{2 }– ch^{2}
Now we have x = t^{2} + t + 1 and y = t^{2} – t + 1
(Adding and subtracting values of x and y)
Eliminating t, 2 (x + y) = (x – y)^{2 }+ 4 ......... (1)
⇒ x^{2} – 2xy + y^{2} – 2x – 2y + 4 = 0 ......... (2)
Here, a = 1, h = – 1, b = 1, g = – 1, f = – 1, c = 4
∴Δ ≠ 0. and h^{2} = ab
Hence the given curve represents a parabola.
If x + y = k is normal to y^{2 }= 12 x, then k is (2000S)
y = mx + c is normal to the parabola y^{2} = 4 ax if c = – 2am – am^{3}
Here m = –1, c = k and a = 3
∴ c = k = – 2 (3) (–1) – 3 (–1)^{3} = 9
If the line x – 1 = 0 is the directrix of the parabola y^{2} – kx + 8 = 0, then one of the values of k is (2000S)
KEY CONCEPT : The directrix of the parabola y^{2} = 4a (x – x_{1}) is given by x = x_{1} – a.
y^{2} = kx – 8 ⇒ y^{2} =
Directrix of parabola is x =
Now, x = 1 also coincides with =
On comparision, =1, or k^{2}  4k  32 = 0
On solving we get k = 4
The equation of the common tangent touching the circle (x 3)^{2} + y^{2 }= 9 and the parabola y^{2 }= 4x above the xaxis is (2001S)
Let the equation of tangent to y^{2} = 4x be y =
where m is the slope of the tangent.
If it is tangent to the circle (x – 3)^{2} + y^{2} = 9 then length of perpendicular to tangent from centre (3, 0) should be equal to the radius 3.
∴ Tangents are x –y + 3=0 and
x +y+ 3 = 0 out of which x –y + 3=0 meets the parabola at (3, 2) i.e., above xaxis.
The equation of the directrix of the parabola y^{2} + 4y + 4x + 2 = 0 is (2001S)
y^{2} + 4y + 4x + 2 = 0
y^{2} + 4y + 4 = – 4x + 2 (y + 2)^{2} = –4 (x – 1/2)
It is of the form Y^{2} = – 4AX
whose directrix is given by X = A
∴ Req. equation is x – 1/2 = 1 ⇒ x = 3/2.
If a > 2b > 0 then the positive value of m for which is a common tangent to x^{2} + y^{2} = b^{2} and (x – a)^{2} + y^{2} = b^{2} is (2002S)
Given that a > 2b > 0 and m > 0
Also ...(1)
is tangent to x^{2} + y^{2} = b^{2} ...(2)
as well as to (x – a)^{2} + y^{2} = b^{2} ...(3)
∵ (1) is tangent to (3)
[length of perpendicular from (a, 0) to (1) = radius b]
or am = 0 (not possible as a, m > 0)
(∵ m > 0)
The locus of the midpoint of the line segment joining the focus to a moving point on the parabola y^{2} = 4ax is another parabola with directrix (2002S)
If (h, k) is the mid point of line joining focus (a, 0) and Q (at^{2}, 2at) on parabola then k = at
Eliminating t, we get 2h =
⇒ k^{2} = a (2h – a) ⇒ k^{2} = 2a (h – a/2)
∴ Locus of (h, k) is y^{2} = 2a (x – a/2)
whose directrix is (x – a/2) =
⇒ x = 0
The equation of the common tangent to the curves y^{2} = 8x and xy = –1 is (2002S)
The given curves are y^{2} = 8x ...(1)
and xy = – 1 ...(2)
If m is the slope of tangent to (1), then eqn of tangent is y = mx + 2/m
If this tangent is also a tangent to (2), then putting value of y in curve (2)
= 0 ⇒ m2 x2 + 2x + m = 0
We should get repeated roots for the eqn (condition of tangency)
⇒ D = 0
∴ (2)^{2} – 4m^{2}.m = 0 ⇒ m^{3} = 1 ⇒ m = 1
Hence required tangent is y = x + 2
The area of the quadrilateral formed by the tangents at the end points of latus rectum to the ellipse is
The given ellipse is
Then a2 = 9, b2 = 5 ⇒ e
∴ end point of latus rectum in first quadrant 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) = = 27 sq. units.
The focal chord to y^{2} = 16x is tangent to (x – 6)^{2} + y^{2 }= 2, then the possible values of the slope of this chord, are (2003S)
For parabola y^{2} = 16x, focus ≡ (4, 0).
Let m be the slope of focal chord then eq^{n} is y = m (x – 4) ...(1)
But given that above is a tangent to the circle (x – 6)^{2 }+ y^{2} = 2
With Centre, C (6, 0), r = 2
∴ Length of ⊥^{ lar }from (6, 0) to (1) = r
⇒ 2m^{2} = m^{2} + 1 ⇒ m^{2} = 1 ⇒ m = ±1
For hyperbola which of the following remains constant with change in ‘α’ (2003S)
The given eq^{n} of hyperbola is
⇒ a = cosα, b = sinα
⇒ ae = 1
∴ foci ( + 1 , 0)
∴ foci remain constant with respect to α.
If tangents are drawn to the ellipse x^{2 }+ 2y^{2} = 2, then the locus of the midpoint of the intercept made by the tangents between the coordinate axes is (2004S)
Any tangent to ellipse
⇒ 2h = 2 secθ and 2k = cosecθ (Using mid pt. formula)
Required locus,
The angle between the tangents drawn from the point (1, 4) to the parabola y^{2} = 4x is (2004S)
y = mx + 1/m
Above tangent passes through (1, 4) ⇒ 4 = m + 1/m ⇒ m^{2} – 4m + 1 = 0
Now angle between the lines is given by
If the line = 2 touches the hyperbola x^{2} – 2y^{2 }= 4, then the point of contact is (2004S)
Equation of tangent to hyperbola x^{2 }– 2y^{2} = 4 at any point (x_{1}, y_{1}) is xx_{1 }– 2yy_{1} = 4
Comparing with 2x +y = 2 or 4 x + 2y=4
⇒ x_{1} = 4 and – 2y_{1} = 2 ⇒ (4, – ) is the required point.
The minimum area of triangle formed by the tangent to the & coordinate axes is (2005S)
Any tangent to the ellipse
at P (a cosθ, b sinθ) is
It meets coordinate axes at A (a secθ, 0) and B (0, b cosecθ)
∴ Area of ΔOAB = x a secθ × b cosecθ
For Δ to be min, sin ^{2}θ should be max. and we know max value of sin^{ 2}θ= 1
∴ Δ max = ab sq. units.
Tangent to the curve y = x^{2 }+ 6 at a point (1, 7) touches the circle x^{2 }+ y^{2} + 16x + 12y + c = 0 at a point Q. Then the coordinates of Q are (2005S)
The given curve is y = x^{2} + 6 Equation of tangent at (1, 7) is (y + 7) = x .1 + 6
⇒ 2x – y + 5 = 0 ...(1)
As given this tangent (1) touches the circle x^{2} + y^{2} +16x + 12y + c = 0 at Q
Centre of circle = (– 8, – 6).
Then equation of CQ which is perpendicular to (1) and passes through (– 8, – 6) is y + 6 = (x + 8)
⇒ x + 2y + 20 = 0 ...(2)
Now Q is pt. of intersection of (1) and (2)
∴ Solving eq^{n} (1) & (2) we get x = – 6, y = – 7
∴ Req. pt. is (– 6, – 7).
The axis of a parabola is along the line y = x and the distances of its vertex and focus from origin are and respectively. If vertex and focus both lie in the first quadrant, then the equation of the parabola is (2006  3M, –1)
Since, distance of vertex from origin is and focus is
∴ Vertex is (1, 1) and focus is (2, 2), directrix x + y = 0
∴ Equation of parabola is
( x – 2)^{2} + (y – 2)^{2}
⇒ 2(x^{2}  4x+ 4)+ 2(y^{2}  4y + 4) = x^{2} + y^{2}+ 2xy
⇒ x^{2 }+ y^{2} – 2xy = 8 (x + y – 2)
⇒ (x – y)^{2} = 8 (x + y – 2)
A hyperbola, having the transverse axis of length 2 sinθ, is confocal with the ellipse 3x^{2} + 4y^{2} = 12. Then its equation is (2007  3 marks)
The length of transverse axis = 2 sinθ = 2a
⇒ a = sinθ
Also for ellipse 3x^{2} + 4y^{2} = 12
∴ Focus of ellipse =
As hyperbola is confocal with ellipse, focus of hyperbola = (1, 0) ⇒ ae = 1 ⇒ sinθ × e = 1 ⇒ e = cosecθ
∴ b^{2} = a^{2} (e^{2} – 1) = sin^{2}θ (cosec^{2}θ – 1) = cos^{2}θ
∴ Equation of hyperbola is
or, x^{2}cosec^{2}θ – y^{2} sec^{2}θ = 1
Let a and b be nonzero real numbers. Then, the equation (ax^{2} + by^{2} + c) (x^{2} – 5xy + 6y^{2}) = 0 represents (2008)
x^{2} – 5xy + 6y^{2 }= 0 represents a pair of straight lines given by x – 3y = 0 and x – 2y = 0.
Also ax^{2} + by^{2 }+ c = 0 will represent a circle if a = b and c is of sign opposite to that of a.
Consider a branch of the hyperbola with vertex at the point A. Let B be one of the end points of its latus rectum. If C is the focus of the hyperbola nearest to the point A, then the area of the triangle ABC is (2008)
The given hyperbola is
∴ a = 2, b =
Clearly ΔABC is a right triangle.
=
=
The line passing through the extremity A of the major axis and extremity B of the minor axis of the ellipse x^{2} + 9y^{2} = 9 meets its auxiliary circle at the point M. Then the area of the triangle with vertices at A, M and the origin O is (2009)
The given ellipse is x^{2 }+ 9y^{2 }= 9 or
So, that A( 3,0) and B (0,1)
∴ Equation of AB is
or x + 3y – 3 = 0 (1)
Also auxillary circle of given ellipse is x^{2} +y^{2}= 9 (2)
Solving equation (1) and (2), we get the point M where line AB meets the auxillary circle.
Putting x = 3 – 3y from eq^{n} (1) in eq^{n} (2) we get (3  3 y)^{2 }+y^{2 }= 9
⇒ 9 18y + 9y^{2 }+y^{2 }= 9 ⇒ 10y^{2} 18y=0
Clearly M
∴ Area of ΔOAM =
The normal at a point P on the ellipse x^{2} + 4y^{2 }= 16 meets the x  axis at Q. If M is the mid point of the line segment PQ, then the locus of M intersects the latus rectums of the given ellipse at the points (2009)
The given ellipse is
such that a^{2 }= 16 and b^{2} = 4
Let P(4cosθ, 2sinθ) be any point on the ellipse, then equation of normal at P is
4 x sinθ  2y cosθ = 12 sinθ cosθ
∴ Q, the point where normal at P meets x –axis, has coordin ates (3 cosθ, 0)
∴ Mid point of PQ is M
For locus of point M we consider
and y = sinθ
and sinθ= y
...(1)
Also the latus rectum of given ellipse is
Solving equations (1) and (2), we get
∴ The required points are
The locus of the orthocentre of the triangle formed by the lines
(1+ p) x – py + p (1+ p) = 0,
(1+ q) x – qy + q (1+ q) = 0,
and y = 0, where p ≠ q, is (2009)
The triangle is formed by the lines
AB : (1 + p) x  py + p (1 +p)= 0
AC : (1+ q)x qy +q(1+ q)= 0
BC :y =0
So that the vertices are A( pq, ( p + 1)(q+ 1)), B( p, 0), C (q, 0)
Let H (h,k) be the orthocentre of Δ ABC.
Then as AH ^ BC and passes through A( pq, ( p + 1)(q+ 1)) The eq^{n} of AH is x = pq
∴ h = pq ...(1)
Also BH is perpendicular to AC
(using eqn (1)
⇒ k = pq ...(2)
From (1) and (2) we observe h + k = 0
∴ Locus of (h, k) is x + y = 0 which is a straight line.
Let P(6, 3) be a point on the hyperbola If the
normal at the point P intersects the xaxis at (9, 0), then the eccentricity of the hyperbola is (2011)
For hyperbola
, we have
∴ Slope of normal at P (6, 3)
∴ Equation of normal is
As it intersects xaxis at (9, 0)
...(1)
Also for hyperbola, b^{2} = a^{2} (e^{2} – 1)
Using a^{2 }= 2b^{2}; we get b^{2} = 2b^{2} (e^{2} – 1)
or
Let (x, y) be any point on the parabola y^{2} = 4x. Let P be the point that divides the line segment from (0, 0) to (x, y) in the ratio 1 : 3. Then the locus of P is (2011)
Let A (x, y) = (t^{2}, 2 t) be any point on parabola y^{2} = 4x.
Let P (h, k) divides OA in the ratio 1 : 3
Then (h, k) =
∴ locus of P (h, k) is x = y^{2}.
The ellipse is inscribed in a rectangle R whose sides are parallel to the coordinate axes. Another ellipse E_{2} passing through the point (0, 4) circumscribes the rectangle R. The eccentricity of the ellipse E_{2} is (2012)
As rectangle ABCD circumscribed the ellipse
∴ A = (3, 2)
Let the ellipse circumscribing the rectangle ABCD is
Given that it passes through (a, 4)
∴ b^{2} = 16
Also it passes through A (3, 2)
The common tangents to the circle x^{2 }+ y^{2} = 2 and the parabola y^{2} = 8x touch the circle at the points P, Q and the parabola at the points R, S. Then the area of the quadrilateral PQRS is(JEE Adv. 2014)
Let the tangent to y^{2} = 8x be y =
If it is common tangent to parabola and circle, thenis a tangent to x^{2 }+ y^{2 }= 2
⇒ m^{4 }+ m^{2} – 2 = 0 ⇒ (m^{2} + 2) (m^{2} – 1) = 0 ⇒ m = 1 or – 1
∴ Required tangents are y = x + 2 and y = – x – 2
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130 videos359 docs306 tests
