question_from_jee_mains_conic_.pdf

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Mathematics Conic IITJEE-2016Level III (Previous year JEE Mains Question)1. The locus of the foot of perpendicular drawn from the centre of the ellipse 2 23 6x y on any tangent to it, is :

(A) 22 2 2 26 2x y x y (B) 22 2 2 26 2x y x y (C) 22 2 2 26 2x y x y (D) 22 2 2 26 2x y x y

1.(C) Equation of ellipse is 2 23 6x y Equation of the tangent is 1x cos y sin

a b

Let (h, k) be any point on the locus, then

1h kcos sina b

.. (i)

Slope of the tangent line is b cota

.

Slope of perpendicular drawn from centre (0, 0) to (h, k) is kh

.

Since, both the lines are perpendicular.

1k b coth a

cos sin

ha kb

cos ha, sin kb

2. The equation of the circle passing through the foci of the ellipse2 2

116 9x y and having centre at (0, 3) is :

(A) 2 2 6 7 0x y y (B) 2 2 6 7 0x y y (C) 2 2 6 5 0x y y (D) 2 2 6 5 0x y y

2.(A) Given equation of ellipse is2 2

116 9x y

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Hence, 94 3 116

a , b , e 74

Foci is 70 4 0 7 04ae, , , Radius of the circle, 2 2 7 9 16 4r ae b Now, equation of circle is 2 20 3 16x y 2 2 6 7 0x y y

3. Given A circle, 2 22 2 5x y and a parabola, 2 4 5y x .Statement I : An equation of a common tangent to these curves is 5y x .

Statement II : If the line, 5 0y mx , mm

is the common tangent, then m satisfies 4 23 2 0m m .(A) Statement I is true, Statement II is true; Statement II is correct explanation of statement I(B) Statement I is true, Statement II is true; Statement II is not a correct explanation of statement I(C) Statement I is true, Statement II is false(D) Statement I is false, Statement II is true

3.(B) Equation of circle can be rewritten as 2 2 52

x y

Here, centre (0, 0) and radius 52

Let common tangent be 5y mxm

The perpendicular from centre to the tangent is equal to radius.

2

5521

m

m

21 2m m 2 21 2m m 4 2 2 0m m 2 22 1 0m m 1m 2 2 0 asm , m R 5y x Both statements are correct as 1m satisfies the given equation of Statement II but Statement II is not a correctexplanation of statement I

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4. Statement I : An equation of a common tangent to the parabola 2 16 3y x and the ellipse 2 22 4x y is2 2 3y x .

Statement II : If the line 4 3y mxm

, 0m is a common tangent to the parabola 2 16 3y and theellipse 2 22 4x y , then m satisfies 4 22 24m m .(A) Statement I is true, Statement II is true; Statement II is correct explanation of statement I(B) Statement I is true, Statement II is true; Statement II is not a correct explanation of statement I(C) Statement I is true, Statement II is false(D) Statement I is false, Statement II is true

4.(C) Statement I : Analysis given, a parabola 2 16 3y x and an ellipse 2 22 4x y To find the equation of common tangent to the given parabola and the ellipse. This can be very easily done bycomparing the standard equation of tangents. Standard equation of tangent with slope m to the parabola

2 16 3y x is4 3y mxm

.. (i)

Standard equation of tangent with slope m to the ellipse2 2

12 4x y is

22 4y mx m If a line L is a common tangent to the both parabola and ellipse, thenL should be tangent to parabola; i.e., its equation should be like equation (i).L should be tangent to ellipse; i.e., its equation should be like equation (ii).i.e., L must be like both of the equations (i) and (ii).

Hence, comparing equations (i) and (ii), we get 24 3 2 4mm

On squaring both sides we get

2 22 4 48m m 4 22 24 0m m 2 26 4 0m m 2 4m 2 6m 2m Substituting 2m in the equation (i), we get the required equation of the common tangents as

2 2 3 and 2 2 3y x y x Hence, Statement I is correct.

Statement II : In the Statement II, we have already seen that, if the line 4 3y mxm

is a common tangent to

the parabola 2 16 3y x and the ellipse2 2

12 4x y , then it satisfies the equation 4 22 24 0m m . Hence,

Statement II is also correct but is not able to explain the Statement I. It is an intermediate step in the final answer.

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5. An ellipse is drawn by taking a diameter of the circle 2 21 1x y as its semi-minor axis and a diameter of thecircle 22 2 4x y as semi-major axis. If the centre of the ellipse is at the origin and its axes are the coordinateaxes, then the equation of the ellipse is :

(A) 2 24 4x y (B) 2 24 8x y (C) 2 24 8x y (D) 2 24 16x y 5.(D) Given,

(i) An ellipse whose semi-minor axis coincides with one of the diameters of the circle 2 21 1x y .(ii) The semi-major axis of the ellipse coincides with one of the diameter of circle 22 2 4x y .(iii) The centre of the ellipse is at origin.(iv) The axes of the ellipse are coordinate axes.To find the equation of the ellipse.

Diameter of circle 2 21 1x y is 2 units and that of circle 22 2 4x y is 4 units. Semi-major axis of ellipse, 2b units and semi-major axis of ellipse, 4a units.Hence, the equation of the ellipse is

2 2

2 2 1x ya b

2 2

116 4x y

2 24 16x y

6. Equation of the ellipse whose axes are the axes of coordinates and which passes through the point 3 1, and haseccentricity 2

5is :

(A) 2 25 3 48 0x y (B) 2 23 5 15 0x y (C) 2 25 3 32 0x y (D) 2 23 5 32 0x y

6.(D) Given,2 2

2 2 1x ya b

Passes through 3 1P , and 25

e

2

22 2 2

9 1 1 and 1 bea a a

2

2 2 29 5 21 and 1

53b

a a a

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2

2 227 5 31

53b

anda a

2 232 32and3 5

a b Equation on ellipse

2 23 5 132 32x y 2 23 5 32x y

7. If two tangents drawn from a point P to the parabola 2 4y x are at right angles, then the locus of P is :(A) x = 1 (B) 2 1 0x (C) 1x (D) 2 1 0x

7.(C) We know that, the tool of point P from which two perpendicular tangents are drawn to the parabola, is the directrixof the parabola.Hence, the required locus is 1x

8. The ellipse 2 24 4x y is inscribed in a rectangle aligned with the coordinate axes which in turn is inscribed inanother ellipse that passes through the point (4, 0). Then, the equation of the ellipse is :(A) 2 212 16x y (B) 2 24 48 48x y (C) 2 24 64 48x y (D) 2 216 16x y

8.(A) Let the equation of the required ellipse be2 2

2 116x y

b

But the ellipse passes through the point (2, 1).

21 1 14 b 2

1 34b

2 43

b Hence, equation is

2 23 116 4x y 2 212 16x y

9. Parabola has the origin as its focus and the line 2x as the directrix. Then, the vertex of the parabola is at :(A) (2, 0) (B) (0, 2) (C) (1, 0) (D) (0, 1)

9.(A) Since, the vertex is the mid-point of the focus and foot of the directrix vertex of the parabola at (1, 0).

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10. A focus of an ellipse is at the origin. The directrix is the line 4x and the eccentricity is 12

, then the length of

semi-major axis is :(A) 5

3(B) 8

3(C) 2

3(D) 4

3

10.(B) 14 and2

aae e

e

2 42a

a

3 42a 8

3a

11. For the hyperbola2 2

2 2 1x y

cos sin , which of the following remains constant when varies ?

(A) Eccentricity (B) Directrix(C) Absciddae of vertices (D) Abscissae of foci

11.(B) The given equation of hyperbola is2 2

2 2 1x y

cos sin

Here, 2 2a cos and 2 2b sin

Now,2

21b

ea

2

221 1

sine tan

cos

e sec

Coordinates of foci are 0ae, i.e., 1 0, .Hence, abscissae of foci remains constant when varies.

12. In an ellipse, the distances between its foci is 6 and minor axis is 8. Then, its eccentricity is :

(A) 12

(B) 45

(C) 15

(D) 35

12.(D) Given, that, 2 6 and 2 8ae b

3 and 4ae b 34

ae

b

2 2

216

9b ea

We know that,

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22

2 1b

ea

2

216 19e

e

216 9 19

e

2 925

e

35

e

13. The locus of the vertices of the family of parabolas3 2 2

23 2

a x a xy a is :

(A) 34

xy (B) 3516

xy (C) 64105

xy (D) 10564

xy

13.(D) The given equation of parabola is3 2 2

23 2

a x a xy a

3

2 323 2

ay a x xa

32

2 23 9 92

3 2 16 16ay a x x

a a a

3 3

23 92

3 4 316a ay a x

a a

233 32

16 3 4a ay a x

a

2335 3

16 3 4a ay x

a

Thus, the vertices of parabola is 3 35

4 16a

,

a

Let, 3 35and

4 16ah k

a

Now, 10564

hk

Thus, the locus of vertices of a parabola is 10564

xy

14. Area of the greatest rectangle that can be inscribed in the ellipse2 2

2 2 1x ya b

is :

(A) a sq unitb

(B) ab sq unit (C) ab sq unit (D) 2ab sq unit

14.(D) Let the coordinates of the vertices of rectangle ABCD be A a cos , b sin , B a cos , b sin , C a cos , b sin and D a cos , b sin

Then length of rectangle, 2AB a cos and breadth of rectangle, 2AD b sin Area of rectangle, 2 2A AB AD a cos b sin 2A ab sin .. (i)

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On differentiating equation (i) w.r.t. , we get

2 2 2dA ab cosd

For maxima or minma, put 0dAd

4 2 0ab cos 22

4

Now,2

2 8 2d A

ab sind

At,2

2 04d A

,

d

Area is maximum at

4

Maximum area of rectangle 2 ab sq units [from equation (i)]Alternate Solution from equation (i),

Area of rectangle, 2 2A ab sin A is maximum when 2 1sin Maximum area of rectangle 2ab squnits

15. The locus of a point P , moving under the condition that the line y x is a tangent to the hyperbola2 2

2 2 1x ya b

, is :(A) A hyperbola (B) A parabola (C) A circle (D) An ellipse

15.(A) Since, the line y x is tangent to the hyperbola2 2

2 2 1x ya b

2 2 2 2a b So, locus of , is 2 2 2 2y a x b 2 2 2 2 0a x y b Since, this equation represents a hyperbola, so locus of a point P , is a hyperbola.

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16. If the pair of lines 2 22 0ax a b xy by lie along diameters of a circle and divide the circle into four sectorssuch that the area of one of the sectors is thrice the area of another sector, then

(A) 2 23 2 3 0a ab b (B) 2 23 10 3 0a ab b (C) 2 23 2 3 0a ab b (D) 2 23 10 3 0a ab b

16.(A) Given equation of pair of lines is 2 22 0ax a b xy by Here, H a b, A a, B b Since, 4

4

Angle between lines is given by22 H AB

tanA B

22

14

a b abtan

a b 2 2 2 22 4a b ab a b ab

2 23 3 2 0a b ab

17. If the sum of the slopes of the lines given by 2 22 7 0x cxy y is four times their product, then c has the value :(A) 1 (B) 1 (C) 2 (D) 2

17.(C) The given pair of lines is 2 22 7 0x cxy y On comparing the standard equation 2 22 0ax hxy by , we get 1 2 2 7a , h c, b Now, 1 2

2 27

h cm m

b and 1 2 17

am m

b

According to the given condition, 1 2 1 24m m m m 2 4

7 7c 2c

18. If 0a and the line 2 3 4 0bx cy d passes through the point of intersection of the parabolas2 24 and 4y ax x ay , then :

(A) 22 2 3 0d b c (B) 22 3 2 0d b c (C) 22 2 3 0d b c (D) 22 3 2 0d b c

18.(A) Given equation of parabola are 2 24 and 4y ax x ay The point of intersection of parabolas are 0 0A , and 4 4B a, aAlso, given line 2 3 4 0bx cy d passes through the point A and B, respectively. d = 0 (i)

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and 2 4 3 4 4 4 0b a c a a d 2 3 0ab ac d 2 3 0a b c 0d 2 3 0b c (ii)On squaring and adding equations (i) and (ii), we get

22 2 3 0d b c

19. The eccentricity of an ellipse with its centre at the origin, is 12

. If one of the directrices is x = 4, then the equation of

the ellipse is :

(A) 2 23 4 1x y (B) 2 23 4 12x y (C) 2 24 3 12x y (D) 2 24 3 1x y 19.(B) Since, equation of directrix is x = 4, then major axis of an ellipse is long X-axis.

4ae 14

2a 1

2e

2a Now, 2 2 21b a e 2 1 34 1 4

4 4b

2 3b Hence, equation of ellipse is

2 22 21 or 3 4 12

4 3x y

x y

20. The normal at the point 21 12bt , bt on a parabola meets the parabola again in the point 22 22bt , bt , then :(A) 2 1

1

2t t

t (B) 2 1

1

2t t

t (C) 2 1

1

2t t

t (D) 2 1

1

2t t

t

20.(A) Equation of the normal at point 21 12bt , bt on parabola is 31 1 12y t x bt bt It also passes through 22 22bt , bt , then

2 32 1 2 1 12 2bt t bt bt bt

2 22 1 2 1 1 2 1 2 12 2t t t t t t t t t t 1 2 12 t t t 2 1

1

2t t

t

21. The foci of the ellipse2 2

2 116x y

b and the hyperbola

2 2 1144 81 25x y coincide. Then, the value of 2b is :

(A) 1 (B) 5 (C) 7 (D) 9

21.(C) Given equation of hyperbola can be rewritten as2 2

2 2 112 95 5

x y

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Eccentricity,2

221

be

a

2 9 251

16 16e

54

e The foci of a hyperbola are

12 50 0 3 05 4

a e , , ,

Given equation of ellipse is2 2

2 116x y

b

Foci of an ellipse are 0 4 0ae, e, but given focus of ellipse and hyperbola coincide, then4 3e 3

4e

Also, 22 2 91 16 1 169 9 716

b a e 22. The equation of the directrix of the parabola 2 4 4 2 0y y x is :

(A) 1x (B) 1x (C) 3 2x / (D) 3 2x /

22.(D) Given equation of parabola can be rewritten as 2 12 42

y x Let 12 and

2y Y x X

2 4y X Here, 1a Equation of directrix is X = a 1 1

2x 3

2x

23. The radius of the circle passing through the foci of the ellipse2 2

116 9x y and having its centre at (0, 3) is :

(A) 4 units (B) 3 units (C) 12 units (D) 72

units

23.(A) The equation of an ellipse is2 2

116 9x y

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Here, 4 3a , b

Eccentricity,2

29 71 1

16 4b

ea

Foci of an ellipse are 40 07

ae, i.e., ,

Radius of required circle

2 27 0 0 3 7 9 16 4 units

question_from_jee_mains_conic_.pdf

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