JEE Advanced (Subjective Type Questions): Conic Sections | Chapter-wise Tests for JEE Main & Advanced PDF Download

Q.1. Suppose that the normals drawn at three different points on the parabola y² = 4x  pass through the point (h, k). Show that h > 2. (1981 - 4 Marks) 

Ans.  2

Sol. The equation of a normal to the parabola y² = 4ax in its slope form is given by 

y = mx – 2am – am³
∴Eq. of normal to y² = 4x, is y = mx – 2m – m³ ...(1)

Since the normal drawn at three different points on the parabola pass through (h, k), it must satisfy the equation (1)
∴k = mh – 2m – m³ ⇒ m³ – (h – 2) m + k = 0
This cubic eq. in m has three different roots say m₁, m₂, m₃
∴m₁ + m₂ + m₃ = 0 ...(2)
m₁m₂ + m₂m₃ + m₃m₁ = – (h – 2) ...(3)
Now, (m₁ + m₂ + m₃)² = 0 [Squaring (2)]
⇒ m₁² + m₂²+m₃² = – 2 (m₁m₂ + m₂m₃ + m₃m₁) ⇒ m₁² + m₂² + m₃² = 2 (h – 2) [Using (3)]
Since LHS of this equation is he sum of perfect squares, therefore it is + ve
∴h – 2 > 0 ⇒ h > 2 Proved

Q.2. A  is a point on the parabola y² = 4ax. The normal at A cuts the parabola again at point B. If AB subtends a right angle at the vertex of the parabola. find the slope of AB. (1982 - 5 Marks) 

Ans. m =±

Sol. Parabola y² = 4ax.
Let at any pt A equation of normal is y = mx – 2am – am³. ...(1)
Combined equation of OA and OB can be obtained by making equation of parabola homogeneous with the help of normal.
∴Combined eq. of OA and OB is

[ From eqn. (1) using]

⇒ 4mx² – 4xy – (2m + m³) y² = 0 But angle between the lines represented by this pair is 90°.
⇒ coeff. of x² + coeff of y² = 0 ⇒  4m – 2m – m³ = 0

⇒ m³ – 2m = 0 ⇒ m = 0,  , –
But for m = 0 eq. of normal becomes y = 0 which does not intersect the parabola at any other point.
∴ m =±

Q.3. Three normals are drawn from the point (c, 0) to the curve y² = x. Show that c must be greater than 1/2. One normal is always the x-axis. Find c for which the other two normals are perpendicular to each other. (1991 -  4 Marks) 

Ans. c = 3/4

Sol. Given parabola is y² = x.

Normal is y =

As per question  this normal passes through (c, 0) therefore, we get

...(1)

m = 0 shows normal is y = 0 i.e. x-axis is always a normal.

At c =         from (1)  m = 0

∴for other real values of m, c > 1/2

Now for other two normals to be perpendicular to each other, we must have m₁.m₂ = – 1
Or in other words, if m₁,m₂ are roots of   = 0, then product of roots = – 1

⇒ c = 3/4

Q.4. Through the vertex O of parabola y² = 4x, chords OP and OQ are drawn at right angles to one another. Show that for all positions of P, PQ cuts the axis of the parabola at a fixed point. Also find the locus of the middle point of PQ. (1994 -  4 Marks)

Ans. y² = 2 (x – 4)

Sol. Let the equation of chord OP be y = mx.
Then eqn of chord OQ will be   [∵ OQ ⊥ OP]
P is pt. of intersection of y = mx and y² = 4x.

Solving the two we get 

Q is pt. of intersection of and y² = 4x.

Solving the two we get Q (4m², – 4m)

Now eq. of PQ is

⇒ (1 – m²) y + 4m – 4m³ = mx – 4m³ ⇒ mx – (1 – m²) y – 4m = 0
This line meets x-axis where y = 0 i.e. x = 4
⇒ OL = 4, which is constant as independent of m. Again let (h, k) be the mid pt. of PQ, then

⇒ 2h = k² + 8    ⇒ k² = 2 (h – 4)
∴ Locus of (h, k) is y² = 2 (x – 4)

Q.5. Show that the locus of a point that divides a chord of slope 2 of the parabola y² = 4x internally in the ratio 1: 2 is a parabola. Find the vertex of this parabola.   (1995 -  5 Marks) 

Ans. (2/9, 8/9)

Sol. Let P (t₁² , 2t₁) and Q (t₂² , 2t₂) be the ends of the chord PQ of the parabola

y² = 4x   ...(1)
∴Slope of chord PQ

⇒ t₂ + t₁ = 1 ...(2)

If R(x₁, y₁) is a point dividing PQ internally in the ratio 1 : 2, then

⇒ t₂² + 2t₁² =3x₁           ...(3)
and t₂ + 2t₁= (3y₁) /2      ...(4)
From (2) and (4), we get

Substituting in (3), we get

⇒ (9/4)y₁² – 4y₁ =x₁-2

∴Locus of the point R ( x₁ ,y₁) is (y – 8/9)² = (4/9) (x – 2/9) which is a parabola having vertex at the point (2/9, 8/9).

Q.6. Let ‘d’ be the perpendicular distance from the centre of the ellipse  to the tangent drawn at a point P on the ellipse. If F₁ and F₂ are the two foci of the ellipse, then show that 
 (1995 -  5 Marks) 

Ans. 

Sol. Equation to the tangent at the point P (a cosθ, b sinθ) on x²/a² + y²/b² = 1 is

...(1)

∴d = perpendicular distance of (1) from the centre (0, 0) of the ellipse

=

=  4 (a² – b²) cos²θ = 4a² e² cos²θ....(2)
The coordinates of focii F₁ and F₂ are F₁ = (ae, 0) and F₂ = (–ae, 0)

= a (1 – e cosθ) Similarly, PF₂ = a (1 + e cosθ)

∴(PF₁ – PF₂)² = 4a² e² cos²θ ...(3)

Hence from (2) and (3), we have

(PF₁ – PF₂)² =  

Q.7. Points A, B and C lie on the parabola y² = 4ax. The tangents to the parabola at A, B and C, taken in pairs, intersect at points P, Q and R. Determine the ratio of the areas of the triangles ABC and PQR. (1996 - 3 Marks) 

Ans. 2 : 1

Sol. Let the three points on the parabola y² = 4ax be A (at₁² , 2at₁) , B(at₂² , 2at₂) and C (at₃² , 2at₃) .
Then using the fact that equation of tangent to y² = 4ax at (at², 2at) is    , we get equations of tangents at A, B and C as folllows

...(1)

...(2)

...(3)
Solving the above equations pair wise we get the pts.

P (at₁t₂, a (t₁ + t₂))
Q (at₂t₃, a (t₂ + t₃))
R (at₃t₁, a (t₃ + t₁))

Now, area of ΔABC =

...(4)

Also area of ΔPQR 

Expanding along C₁,

...(5)

From equations (4) and (5), we get

∴The required ratio is 2 : 1

Q.8. From a point A common tangents are drawn to the circle x² + y² = a²/2 and  parabola y² = 4ax. Find the area of the quadrilateral formed by the common tangents, the chord of contact of the circle and the chord of contact of the parabola. (1996 - 2 Marks)

Ans. 

Sol. This line will touch the circle x² + y² = a²/2

if          

⇒ 2 = m⁴ + m² ⇒ m⁴ + m² –  2 = 0 

⇒ (m² + 2) (m2 – 1) = 0   ⇒ m = 1, – 1
Thus the two tangents (common one) are y = x + a and y = – x – a
These two intersect each other at (– a, 0) The chord of contact at A (– a, 0) for the circle x² + y² = a²/2 is  
(– a. x) + 0.y = a²/2 i.e., x = – a/2 and the chord of contact at A (– a,0) for the parabola y² = 4ax is 0.y = 2a (x – a) i.e., x = a
Note that DE is latus recturn of parabola y² = 4ax, therefore its lengths is 4a.
Chords of contact are clearly parallel to each other, so req. quadrilateral is a trapezium.

Ar (trap BCDE) = BC + DEx KL

Q.9. A tangent to the ellipse x² + 4y² = 4 meets the ellipse x² + 2y² = 6 at P and Q. Prove that the tangents at P and Q of the ellipse x² + 2y² = 6 are at right angles.(1997 - 5 Marks) 

Ans. 

Sol. The given ellipses are

...(1)

and ...(2)

Then the equation of tangent to (1) at any point T (2 cosθ, sinθ) is given by

or ...(3)
Let this tangent meet the ellipse (2) at P and Q.
Let the tangents drawn to ellipse (2) at P and Q meet each other at R (x₁, y₁)
Then PQ is chord of contact of ellipse (2) with respect to the pt R (x₁, y₁) and is given by

...(4)
Clearly equations (3) and (4) represent the same lines and hence should be identical. Therefore comparing the cofficients, we get

⇒ x₁ = 3 cosθ , y₁ = 3 sinθ ⇒ x₁² +y₁²= 9
⇒ Locus of (x₁, y₁) is x² + y² = 9

 which is the director circle of the ellipse = and
Thus tangents at P and Q are at right ∠’s.
KEY CONCEPT : We know that the director circle is the locus of intersection point of the tangents which are at right ∠.

Q.10. The angle between a pair of tangents drawn from a point P to the parabola y² =4ax is 45°. Show that the locus of the point P is a hyperbola. (1998 - 8 Marks)

Ans. 

Sol. Let P (e, f) be any point on the locus. Equation of pair of tangents from P (e, f) to the parabola y² = 4ax is [ fy – 2a (x + e)]² 
= (f ² – 4ae) (y² – 4ax)    [T² = SS₁]
Here, a = coefficient of x² = 4a² ...(1)
2h = coefficient of xy = – 4af ...(2)
and b = coefficient of y² = f ² – (f ² – 4ae) = 4ae ...(3)
If they include an angle 45°, then

1 tan 45°

or,(a + b)² = 4 (h² – ab)
or, (4a² + 4ae)² = 4 [4a²f ² – (4a²) (4ae)]
or,(a + e)² = f² – 4ae or e² + 6ae + a² – f² = 0
or(e + 3a)² – f ² = 8a² 
Hence the required locus is (x + 3a)² – y² = 8a², which is a hyperbola.

Q.11. Consider the family of circles x² + y² = r², 2 < r < 5. If in the first quadrant, the common taingent to a circle of this family and the ellipse 4x² + 25y² = 100 meets the co-ordinate axes at A and B, then find the equation of the locus of the mid-point of AB. (1999 - 10 Marks) 

Ans. Sol. Let any point P on ellipse 4x² + 25y² = 100 be (5 cosθ, 2 sinθ). So equation of tangent to the ellipse at P will be

Tangent (1) also touches the circle x² + y²  = r², so its distance from origin must be r.
Tangent (2) intersects the coordinate axes at  and  respectively. Let M (h, k) be the midpoint of line segment AB. Then by mid point formula

 ⇒

⇒

Hence locus of M (h, k) is 

Locus is independent of r.

Q.12. Find the  co-ordinates of all the points P on the ellipse  , for which the area of the triangle PON is maximum, where O denotes the origin and N, the foot of the perpendicular from O to the tangent at P.(1999 - 10 Marks) 

Ans. 

Sol. The ellipse is         ..(1)
Since this ellipse is symmetrical in all four quadrants, either there exists no such P or four points, one in each qudrant.
Without loss of generality  we can assume that a > b and P lies in first quadrant.

Let P (a cosθ, b sinθ) then equation of tangent is

Equation of ON is 

Equation of normal at P is ax secθ – by cosecθ = a–²  b²

and NP = OL

∴Z = Area of OPN =x ON x NP

Let u= a² tanθ + b² cotθ

⇒ tanθ = b/a

  u is minimum at θ = tan^–1 b/a

So Z is maximum at θ = tan^–1 b/a

By symmetry, we have four such points

Q.13. Let ABC be an equilateral triangle  inscribed in the circle x² + y² = a². Suppose perpendiculars from A, B, C to the major axis of the ellipse

meets the ellipse respectively, at P, Q, R. so that P, Q, R lie on the same side of the major axis as A, B, C respectively. Prove that the normals to the ellipse drawn at the points P, Q and R are concurrent. (2000 - 7 Marks) 

Ans. 

Sol. Let A, B, C be the point on circle whose coordinates are A = [a cosθ , a sinθ ]

and 

Further, P [ a cosθ , b sinθ ] (Given)

and 

It is given that P, Q, R are on the same side of x-axis as A, B, C.

So required normals to the ellipse are ax secθ– by cosecθ= a² – b² ...(1)

...(2)

...(3)

Now, above three normals are concurrent ⇒ Δ = 0

where Δ 

Multiplying and dividing the different rows R₁, R₂ and R₃ by sinθ cosθ ,

and   respectively, we get

[Operating R₂ → R₂ + R₃ and simplyfing R₂ we get R₂ ≡ R₁]

Q.14. Let C₁ and C₂ be respectively, the parabolas x² = y – 1 and y² = x – 1. Let P be any point on C₁ and Q be any point on C₂. Let P₁ and Q₁ be the reflections of P and Q, respectively, with respect to the line y = x. Prove that P₁ lies on C₂, Q₁ lies on C₁ an d PQ ≥ min{PP₁, QQ₁} . Hence or otherwise determine points P₀ and Q₀ on the parabolas C₁ and C₂ respectively such that P₀Q₀ ≤ PQ for all pairs of points (P,Q) with P on C₁ and Q on C₂. (2000 - 10 Marks)

Ans. Sol. Given that C₁ : x² = y – 1   ; C₂ : y² = x – 1
Let P ( x₁ ,x₁² + 1) on C₁ and Q ( y₂² + 1,y²) on C₂.
Now the reflection of pt P in the line y = x can be obtained by interchanging the values of abscissa and ordiante.
Thus reflection of pt. P ( x₁,x₁² + 1) is P₁ ( x₁² + 1,x₁) and reflection of pt. Q ( y₂² + 1,y²) is  Q₁ ( y² ,y₂² + 1)

It can be seen clearly that P₁ lies on C₂ and Q₁ on C₁.
Now PP₁ and QQ₁ both are perpendicular to mirror line y = x.
Also M is mid pt. of PP₁  (Q P₁ is morror image of P in y = x)

In rt ΔPML,

PL > PM ⇒..(i)
Similarly,

...(ii)

Adding (i) and (ii) we get

⇒ PQ is more than the mean of PP₁ and QQ₁ 
⇒ PQ ≥ min (PP₁, QQ₁) Let min (PP₁, QQ₁) = PP₁

then 

= 2(x₁² + 1 – x₁)² = f (x1)

⇒ f '( x₁) = 4(x₁² + 1 – x₁)(2x₁ – 1)

∴f ' (x₁) = 0 when x₁ = 

Also f ' (x₁) < 0 if x₁

⇒ f (x₁) is min when x₁ = 

Thus if at x₁ = pt P is P₀ on C₁

Similarly Q₀ on C₂ will be image of P₀ with respect to y = x

∴ 

Q.15. Let P be a point on the ellipse  0 < b < a. Let the line parallel to y-axis passing through P meet the circle x² + y² = a² at the point Q such that P and Q are on the same side of x-axis. For two positive real numbers r and s, find the locus of the point R on PQ such that PR : RQ = r : s as P varies over the ellipse. (2001 - 4 Marks) 

Ans. 

Sol. Let the co-ordinates of P be (a cosθ , b sinθ ) then coordinates of Q are (a cosθ , a sinθ )

As R (h, k) divides PQ in the ratio r : s, then

⇒

⇒  ∵ cos²θ + sin²θ = 1

Hence locus of R is   which is equation of an ellipse.

Q.16. Prove that, in an ellipse, the perpendicular from a focus upon any tangent and the line joining the centre of the ellipse to the point of contact meet on the corresponding directrix. (2002 - 5 Marks) 

Ans. Sol. 

Let the ellipse be   and O be the centre.

Tangent at P (x₁, y₁) is = 0 whose 

slope = Focus is S (ae, 0).

Equation of the line perpendicular to tangent at P is

....(1)
Equation of OP is ...(2)
(1) and (2) intersect

⇒ x (a² – b²) = a³e ⇒ x. a²e² = a³e
⇒ x = a/e
Which is the corresponding directrix.

Q.17. Normals are drawn from the point P with slopes m₁, m₂, m₃ to the parabola y² = 4x. If locus of P with m₁ m₂ = a  is a part of the parabola itself then find a. (2003 - 4 Marks) 

Ans. α = 2 

Sol. Let P be the pt. (h, k). Then eqn of normal to parabola y² = 4x from point (h, k), if m is the slope of normal, is y = mx – 2m – m³
As it passes through (h, k), therefore mh – k – 2m – m³ = 0 or, m³ + (2 – h)  + k = 0 ...(1)
Whcih is cubic in m, giving three values of m say m₁, m₂ and m₃.
Then m₁m₂m₃ = – k (from eqn) but given that m₁m₂ = a
∴We get m₃

But m³ must satisfy eqⁿ (1)

⇒ k² – 2α² – hα² – α3 = 0
∴Locus of P (h, k) is y² = α²x + (α³ – 2α²)
But ATQ, locus of P is a part of parabola y² = 4x, therefore comparing the two, we get α² = 4 and α³ – 2α² = 0 ⇒ α = 2

Q.18. Tangent is drawn to parabola y² – 2y – 4x + 5 = 0 at a point P which cuts the directrix at the point Q. A point R is such that it divides QP externally in the ratio 1/2 : 1. Find the locus of point R. (2004 - 4 Marks) 

Ans. ( x - 1)(y - 1)² + 4=0

Sol. The given eqⁿ of parabola is y² – 2y – 4x + 5 = 0 ...(1)
⇒ (y – 1)² = 4 (x – 1)
Any parametric point on this parabola is P (t² + 1, 2t + 1)
Differentiating (1) w.r. to x, we get

∴Slope of tangent to (1) at point P (t² + 1, 2t + 1) is

∴Eqn of tangent at P (t² + 1, 2t + 1) is

y – (2t + 1)

⇒ yt – 2t² – t = x – t² = 1
⇒ x – yt + (t² +  t – 1) = 0 ...(2)

Now directrix of given parabola is (x – 1) = – 1 ⇒ x = 0

Tangent (2) meets directix at 

Let pt. R be (h, k)

ATQ, R divides the line joing QP in the ratio :1 i.e., 1 : 2
externally.

⇒ h = – (1 + t²) and

⇒ t² = – 1 – h and

Eliminating t, we get 

⇒ 4 = – (1 – k)² (1 – h) ⇒ (h – 1) (k – 1)² + 4 = 0

∴Locus of R (h, k) is (x – 1) (y – 1)² + 4 = 0

Q.19. Tangents are drawn from any point on the hyperbola

to the circle x² + y² = 9. Find the locus of mid-point of the chord of contact. (2005 - 4 Marks) 

Ans. 

Sol. Any pt on the hyperbola  is (3 secθ , 2 tanθ) Then, equation of chord of contact to the circle x² + y² = 9,
with respect to the pt. (3 secθ, 2 tanθ) is (3 secθ) x + (2 tanθ) y = 9 ...(i)
If (h, k) be the mid point of chord of contact then equation of chord of contact will be hx + ky – 9 = h² + k² – 9 (T = S₁)
or, hx + ky = h² + k² ...(ii)
But equations (i) and (ii) represent the same st. line and hence should be identical, therefore, we get

sec²θ – tan²θ = 1

∴Locus of (h, k) is

Q.20. Find the equation of the common tangent in 1^st quadrant to the circle x² + y² = 16 and the ellipse

  Also find the length of the intercept of the tangent between the coordinate axes. (2005 - 4 Marks)

Ans. 

Sol. Let the common tangent to circle x² + y² = 16 and ellipse x²/25 + y² /4 = 1 be

..(i)

As it is tangent to circle x² + y² = 16, we should have

[Using : length of perpendicular from (0,0 ) to (1) = 4]

⇒ 25m² + 4 = 16m² + 16   ⇒   9m² =12

[Leaving + ve sign to consider tangent in I quadrant] ∴Equation of common tangent is 

This tangent meets the axes at  and

∴Length of intercepted portion of tangent between axes