Test: JEE Main 35 Year PYQs- Conic Sections - Airforce X Y / Indian Navy SSR MCQ

Any tangent to the parabola y² = 8ax is...(i)

If (i) is a tangent to the circle, x² + y² = 2a² then,

⇒ m²(1 + m²) = 2 ⇒ (m² + 2)(m² – 1) = 0 ⇒ m = ± 1.
So from (i), y = ± (x + 2a).

Equation of the normal to a parabola y² = 4bx at point
(bt₁²,2bt₁) is y = – t₁ x + 2bt₁+ bt₁³

As given, it also passes through (bt₂² ,2bt₂) then

 ⇒ t₂ + t₁ =

∴ Foci = ( ±3 , 0)

∴ foci of ellipse = foci of hyperbola

∴ for ellipse ae = 3 but a = 4,

∴  

Then b² = a² (1-e2) ⇒ b² =

Solving equations of parabolas y² = 4ax and x² = 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
⇒ d² + (2b + 3c)² = 0

  e =   Directrix ,

Equation of elhipe is  3x² + 4y² = 12

P = (1, 0)  Q = (h, k)

Such that K² = 8h Let (α, β) be the midpoint of  PQ

2α - 1 = h 2β = k.
(2β) ² = 8(2α - 1)

⇒ β² = 4α - 2

⇒ y² -4x+2=0.

Tangent to the hyperbola is

Given that y = αx + β is the tangent of hyperbola

⇒ m = α and a² m² -b² =β²

∴ a²α² -b² =β²

Locus is a² x² - y² = b² which is hyperbola.

⇒ 2(a²e² + b²)=4a²e² ⇒

Also e² = 1- b² / a² =1-e²

Given parabola is  

∴ Vertex of parabola is 

To find locus of this vertex,

 64xy = 105

which is the required locus.

2ae =6 ⇒ ae = 3 ; 2b =8 ⇒ b = 4

b² = a²(1-e²) ;16 = a²- a²e² ⇒ a² = 16 + 9= 25

⇒ a = 5 

     ∴ m₁ = (2x - 5)(2, 0) =-1 ,

m₂ = (2x - 5)_{(3, 0)}=1 ⇒ m₁m₂ =-1

i.e. the tangents are perpendicular to each other.

Given, equation of hyperbola is

We know that the equation of hyperbola is

Here, a² = cos²α and b² = sin²α

We know that, b² = a² (e² - 1)

⇒ sin²α = cos²α(e² - 1)

⇒ sin²α + cos²α = cos²α.e²

⇒ e² = 1 + tan²α  = sec²α  ⇒ e = secα 

Co-ordinates of foci are (± ae, 0) i.e. ( ± 1, 0)

Hence, abscissae of foci remain constant when α varies.

Parabola y² = 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)

Equation of normal at P(x, y) is Y – y  = 

Coordinate of G at X axis is (X, 0) (let)

⇒ 

∴ Co-ordinate 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 ⇒  ⇒ x² – y² = –2c₁

∴ the curve is a hyperbola

Perpendicular distance of directrix from focus

∴ Semi major axis = 8/3

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 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)

⇒ b² = 4/3 [substituting a² = 16 in eqⁿ (a)]

∴ The required ellipse is   or   x² + 12y² =16

The locus of perpendicular tangents is directrix i.e., x =-1

Let the ellipse be

It passes through (– 3, 1) so

Also, b² =a² (1- 2 / 5) ⇒ 5b² =3a² ...(ii)

Solving (i) and (ii) we get a²  = 

So, the equation of the ellipse is 3x² + 5y² = 32

Given equation of ellipse is 2x² + y² = 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 y² = 4ax is

On comparing (1) and (2), we get

Squaring on both the sides, we get 16 (3) = (2m² + 4) m²
⇒   48 = m² (2m² + 4) ⇒    2m⁴ + 4m² – 48 = 0 ⇒ m⁴ + 2m² –24 = 0
⇒    (m² + 6)(m² – 4) = 0 ⇒ m²= 4 (∵ m² ≠ – 6) ⇒ m = ± 2
⇒ Equation of common tangents are
Thus, statement-1 is true.
Statement-2 is obviously true.

Equation of circle is (x – 1)² + y² =1 ⇒ radius = 1 and diameter = 2
∴ Length of semi-minor axis is 2.
Equation of circle is x² + (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

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
x² + y² – 6y – 7 = 0

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 m² (1 + m²) = 2
⇒ m⁴ + m² – 2 = 0  
⇒ (m² + 2)(m² – 1) = 0
⇒ m = ± 1

both statements are correct as m = ±1 satisfies the given equation of statement-2.

Given eqn of  ellipse can be written as

 a² = 6, b² = 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

(x⁴ + y⁴ + 2x² y²) = a²x²+ b²y²

⇒ ( x² + y²)² = a² x² + b²y²

Let tangent to y² = 4x be y =

Since this is also tangent to x² = – 32y

Now, D = 0

Let P(h, k) divides OQ in the ratio 1 : 3
Let any point Q on x² = 8y is (4t, 2t²).

Then by section formula

⇒ 2k =h²

Required locus of P is x² = 2y

Given curve is x² + 2xy – 3y² = 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 end point of latus rectum of ellipse

in first quadrant is   and the tangent at this point intersects x-axis at and y-axis at (0, a).

The given ellipse is 

Then a² = 9, b² = 5   ⇒

∴ end point of latus rectum in first quadrent 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) = sq. units.

Minimum distance
⇒ perpendicular distance Eqⁿ of normal at p(2t², 4t) y = –tx + 4t + 2t³

It passes through C(0, –6)  ⇒ t³ + 2t + 3 = 0  ⇒ t  = – 1

Centre of new circle = P(2t² , 4t) = P(2,-4)

Radius 

∴ Equation of the circle is

⇒ x² + y² – 4x + 8y + 12 = 0

⇒ 4b² = a²e² ⇒ 4a²(e² -1)= a²e² ⇒ 3e² = 4