Test: Single Correct MCQs: Conic Sections | JEE Advanced

Given that  

As r > 1

∴ 1 – r < 0  and 1 + r > 0

∴ Let  1 – r = – a², 1 + r = b², then we get

which is not possible for any real values of x and y.

(a) x² + 2y² ≤ 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) x² – y² ≥ 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) y² ≤ x represents interior region of parabola in which for any two pts, their mid point also lie inside the region.

We have 2x² + 3y² – 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).

Since 1² + 2² = 5 < 9 and 2² + 1¹ = 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.

The focus of parabola y² = 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 y² = 2px ...(i)
and 4x² + 4y² – 4px – 3p² = 0 ...(ii)
can be obtained by solving (i) and (ii) as follows 4x² + 8px – 4px – 3p² = 0 ⇒ (2x + 3p) (2x – p) = 0

⇒ y² = – 3p² (not possible), p² ⇒ y = + p

∴ Required pts are (p/2, p), (p/2, – p)

For ellipse   = 1, a = 4, b = 3

∴ Foci are 

Centre of circle is at (0, 3) and it passes through

, therefore radius of circle =

KEY CONCEPT : Equation of the normal to the hyperbola  

at the point (a secα, b tanα) is given by ax cosα + by cotα = a² + b²

Normals at 

where   and these pass through (h, k)
∴ ah cosθ + bk cotθ = a² + b² ah sinθ + bk tanθ = a² + b²
Eliminating h, bk (cotθ sinθ – tan cosθ ) = (a² + b²)
(sinθ – cosθ ) or k = – (a² + b²)/b

Chord x = 9 meets x² – y² = 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
or9x² – 8y² – 18x + 9 = 0

KEY CONCEPT
The equation ax² + 2hxy + by² + 2gx + 2fy + c = 0 represents a parabola if Δ ≠ 0 and h² = ab where Δ = abc + 2fgh – af ² – bg² – ch²
Now we have x = t² + t + 1 and  y = t² – t + 1

(Adding and subtracting values of x and y)
Eliminating t,   2 (x + y) = (x – y)² + 4 ......... (1)
⇒ x² – 2xy + y² – 2x – 2y + 4  = 0 ......... (2)
Here, a = 1, h = – 1, b = 1, g = – 1, f = – 1, c = 4
∴Δ ≠ 0. and h² = ab
Hence the given curve represents a parabola.

y = mx + c is normal to the parabola y² = 4 ax if c = – 2am – am³
Here m = –1, c = k and a = 3
∴  c = k = – 2 (3) (–1) – 3 (–1)³ = 9

KEY CONCEPT : The directrix of the parabola y² = 4a (x – x₁) is given by x = x₁ – a.

y² = kx – 8 ⇒ y² =

Directrix of parabola is  x =

Now, x = 1 also coincides with  =

On comparision,  =1,  or k² - 4k - 32 = 0
On solving  we get k = 4

Let the equation of tangent to y² = 4x be y =

where m is the slope of the tangent.
If it is tangent to the circle (x – 3)² + y² = 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 x-axis.

y² + 4y + 4x + 2 = 0
y² + 4y + 4 = – 4x + 2 (y + 2)² = –4 (x – 1/2)
It is of the form Y² = –  4AX
whose directrix is given by X = A
∴ Req. equation is x – 1/2 = 1 ⇒ x = 3/2.

Given that a > 2b > 0 and m > 0
Also ...(1)
is tangent to x² + y² = b² ...(2)
as well as to (x – a)² + y² = b² ...(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)

If (h, k) is the mid point of line joining focus (a, 0) and Q (at², 2at) on parabola then k = at

Eliminating t, we get  2h = 

⇒ k² = a (2h – a) ⇒  k² = 2a (h – a/2)

∴ Locus of (h, k) is y² = 2a (x – a/2)

whose directrix is (x – a/2) =

⇒ x = 0

The given curves are y² = 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 – 4m².m = 0 ⇒ m³ = 1 ⇒ m = 1
Hence required tangent is y = x + 2

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 x-axis at A (9/2, 0) and y-axis at B (0, 3)

∴ Area of ΔOAB = 

By symmetry area of quadrilateral = 4 × (Area ΔOAB) =  = 27 sq. units.

For parabola y² = 16x, focus ≡ (4, 0).
Let m be the slope of focal chord then eqⁿ is y = m (x – 4) ...(1)
But given that above is a tangent to the circle (x – 6)² + y² = 2
With Centre, C (6, 0), r = 2
∴ Length of ⊥ ^lar from (6, 0) to (1) = r

⇒ 2m² = m² + 1 ⇒ m² = 1  ⇒ m = ±1

The given eqⁿ of hyperbola is

⇒ a = cosα, b = sinα

⇒ ae = 1
∴ foci ( + 1 , 0)
∴ foci remain constant with respect to α.

Any tangent to ellipse 

⇒ 2h = 2 secθ and 2k = cosecθ (Using mid pt. formula)

Required locus,

y = mx + 1/m
Above tangent passes through (1, 4) ⇒ 4 = m + 1/m ⇒  m² – 4m + 1 = 0
Now angle between the lines is given by

Equation of tangent to hyperbola x² – 2y² = 4 at any point (x₁, y₁) is xx₁ – 2yy₁ = 4
Comparing with 2x +y = 2 or 4 x + 2y=4
⇒ x₁ = 4 and  – 2y₁ = 2 ⇒ (4, – ) is the required point.

Any tangent to the ellipse 

at P (a cosθ, b sinθ) is

It meets co-ordinate 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 ²θ should be max. and we know max value of sin ²θ= 1

∴ Δ max = ab sq. units.

The given curve is y = x² + 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 x² + y² +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 eqⁿ (1) & (2) we get x = – 6, y = – 7
∴ Req. pt. is (– 6, – 7).

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(x² - 4x+ 4)+ 2(y² - 4y + 4) = x² + y²+ 2xy
⇒ x² + y² – 2xy = 8 (x + y – 2)
⇒ (x – y)² = 8 (x + y – 2)

The length of transverse axis = 2 sinθ = 2a
⇒ a = sinθ
Also for ellipse 3x² + 4y² = 12

∴ Focus of ellipse = 

As hyperbola is confocal with ellipse, focus of hyperbola = (1, 0)  ⇒ ae = 1 ⇒ sinθ × e = 1 ⇒ e = cosecθ

∴ b² = a² (e² – 1) = sin²θ (cosec²θ – 1) = cos²θ

∴ Equation of hyperbola is

or, x²cosec²θ – y² sec²θ = 1

x² – 5xy + 6y² = 0 represents a pair of straight lines given by x – 3y = 0 and x – 2y = 0.
Also ax² + by² + c = 0 will represent a circle if a = b and c is of sign opposite to that of a.

The given hyperbola is

∴ a = 2, b = 

Clearly ΔABC is a right triangle.

=

=

The given ellipse is x² + 9y² = 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 x² +y²= 9 (2)
Solving equation (1) and (2), we get the point M where line AB meets the auxillary circle.
Putting  x = 3 – 3y from eqⁿ (1) in eqⁿ (2) we get (3 - 3 y)² +y² = 9
⇒ 9 -18y + 9y² +y² = 9 ⇒ 10y² -18y=0

Clearly M 

∴ Area of ΔOAM =

The given ellipse is

such that a² = 16 and b² = 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 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 eqⁿ 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.

For hyperbola

, we have

∴ Slope of normal at P (6, 3)

∴ Equation of normal is

As it intersects x-axis at (9, 0)

...(1)
Also for hyperbola, b² = a² (e² – 1)
Using a² = 2b²; we get b² = 2b² (e² – 1)

  or 

Let A (x, y) = (t², 2 t) be any point on parabola y² = 4x.

Let P (h, k) divides OA in the ratio 1 : 3

Then (h, k) =

∴ locus of P (h, k) is x = y².

As rectangle ABCD circumscribed the ellipse

∴ A = (3, 2)

Let the ellipse circumscribing the rectangle ABCD is

Given that it passes through (a, 4)
∴ b² = 16
Also it passes through A (3, 2)

Let the tangent to y² = 8x be  y = 

If it is common tangent to parabola and circle, thenis a tangent to x² + y² = 2

⇒ m⁴ + m² – 2 = 0 ⇒ (m² + 2) (m² – 1) = 0 ⇒ m = 1 or – 1
∴ Required tangents are y = x + 2 and y = – x – 2