ELLIPSE-5

Equations of Normals in different forms

(i) Point form

Let equation of ellipse be x2a2+y2b2=1

Equation of normal is a2xx1b2yy1=a2b2

(ii) Parametric form

Let equation of ellipse be x2a2+y2b2=1

Equation of normal is a xsecϕb y\cosecϕ=a2b2

(acosϕ,bsinϕ) is parametric coordinates of P(x1,y1), i.e. P(acosϕ,bsinϕ)

(iii) Slope form

For an ellipse x2a2+y2b2=1

Equation of normal in terms of slope (m) is

y=mx±m(a2b2)a2+b2m2

Condition of normality when y=mx+c is the normal of x2a2+y2b2=1 is

c2=m2(a2b2)2a2+b2 m2 or c=±m(a2b2)a2+b2 m2

Examples

1. If the normal at the point P(θ) to the ellipse x214+y5=1 intersects it again at the point Q(2θ) then cosθ is equal to

(a) 23

(b) 23

(c) 32

(d) 32

Show Answer

Solution :

The normal at (asinθ,bsinθ) is axcosθbysinθ=a2b2

Here a2=14& b2=5

a=14b=5

14xcosθ5ysinθ=145

14xcosθ5ysinθ=9..(1)

It meets the curve again at θ(2θ)(acos2θ,bsin2θ)

i.e. (14cos2θ,5sin2θ)

equation (1) satisfy this point

14cos2θcosθ5sin2θsinθ=9

14(2cos2θ1)cosθ5(2sinθcosθ)sinθ=9

28cos2θ1410cos2θ=9cosθ

18cos2θ9cosθ14=0

(6cosθ7)(3cosθ+2)=0

cosθ=76>1 not possible (cosθ<1)

cosθ=23

option (b) is correct.

2. The equation of the normal to the ellipse x2a2+y2b2=1 at the end of the latus rectum in the first quadrant is

(a) x+ ey ae3=0

(b) xey+aee3=0

(c) xeyae3=0

(d) None of these

Show Answer

Solution :

The end of the latus rectum in the first quadrantal is (ae,b2a)

Equation of normal at (ae,b2a) is

a2xaeb2yb2=a2b2,a2xx1b2yy1=a2b2axeay=a2b2axeay=ea2eb2=ea2e(a2a2e2)=ea2ea2+a2e3axaeya2e3=0xeyae3=0

Correct option is c

3. The condition that the line xcosα+ysinα=p may be a normal to the ellipse x2a2+y2b2=1 then

(a) (a2b2)2=p2(a2sec2α+b2cosec2α)

(c) (a2+b2)2=p2(a2sec2α+b2cosec2α)

(b) (a2b2)2=p2(a2cosec2α+b2sec2α)

(d) None of these

Show Answer

Solution :

The equation of normal to x2a2+y2 b2=1 at ’ θ ’ is

a2xacosθb2ybsinθ(a2b2)=0.(1)

Given that xcosα+ysinαp=0 is normal to the ellipse comparing (1) & (2) we get

a/cosθcosα=b/sinθsinα=a2b2p or acosθcosα=bsinθsinα=a2b2p or cosθcosαa=sinθsinαb=pa2b2cosθ=ap(a2b2)cosα,sinθ=bp(a2b2)sinα

Squaring and adding we get

cos2θ+sin2θ=a2p2(a2b2)2cos2α+b2p2(a2b2)2sin2α

(a2b2)2=p2(a2sec2α+b2cosc2α)

correct option is a

4. If the normals at P(x1,y1),Q(x2,y2) and R(x3,y3) to the ellipse are concurrent, then

(a) |x2y2x2y2x1y1x1y1x3y3x3y3|=1

(b) |x1y1x1y1x2y2x2y2x3y3x3y3|=0

(c) |x2y2x2y2x3y3x3y3x1y1x1y1|=1

(d) None of these

Show Answer

Solution :

The equations of the normals at P(x1,y1),Q(x2,y2) and R(x3,y3) to the ellipse x2a2+y2 b2=1 are

a2xx1b2yy1=a2b2..(1)

a2xx2b2yy2=a2b2..(2)

a2xx3b2yy3=a2b2..(3)

respectively

These lines are concurrent, if

|a2x1b2y1a2b2a2x2b2y2a2b2a2x3b2y3a2b2|=0

a2b2(a2b2)|1x11y111x21y211x31y31|=0

R1x1y1R1;R2x2y2R2;R3x3y3 we get

|y1x1x1y1y2x2x2y2y3x3x3y3|=0

OR |x1y1x1y1x2y2x2y2x3y3x3y3|=0

your correct option is b

Practice questions

1. In the normal at the end of latus rectum of the ellipse x2a2+y2b2=1 with eccentricity e, passes through one end of the minor axis, then :

(a) e2(1+e2)=0

(b) e2(1+e2)=1

(c) e2(1+e2)=1

(d) e2(1+e2)=2

Show Answer Answer: (b)

2. If the normals to x2a2+y2 b2=1 at the ends of the chords 1x+m1y=1 and 2x+m2y=1 are concurrent, then :

(a) a212+b2 m1 m2=1

(b) a212+b2 m1 m2=1

(c) a212b2 m1 m2=1

(d) None of these

Show Answer Answer: (b)

3. If the normal at an end of a latus rectum of an ellipse passes through one extremity of the minor axis, then the eccentricity of the ellipse is given by

(a) e4+e21=0

(b) e4+e25=0

(c) e3=5

(d) None of these

Show Answer Answer: (a)

4. The number of normals that can be drawn from a point to a given ellipse is

(a) 2

(b) 3

(c) 4

(d) 1

Show Answer Answer: (c)

5. If the normal at any point P on the ellipse x2a2+y2 b2=1 meets the axes in G and g respectively, then PG:Pg is equal to

(a) a:b

(b) a2:b2

(c) b2:a2

(d) b:a

Show Answer Answer: (c)

6. If normal to ellipse x2a2+y2b2=1 at (ae,b2a) is passing through (0,2b), then value of eccentricity is

(a) 21

(b) 2(21)

(c) 2(21)

(d) None of these

Show Answer Answer: (c)

7. If normal at any point P to the ellipse x2a2+y2 b2=1,a>b meet the axes at M and N so that PMPN=23, then value of eccentricity e is

(a) 12

(b) 23

(c) 13

(d) None of these

Show Answer Answer: (c)

8. If the tangent drown at point (t2,2t) on the parabola y2=4x is same as the normal drawn at point (5cosθ,2sinθ) on the ellipse 4x2+5y2=20. Then the values of tandθ are

(a) θ=cos1(15)&t=15

(b) θ=cos1(15)&t=15

(c) θ=cos1(25)&t=25

(d) None of these

Show Answer Answer: (a)

9. The normals at four points on the ellipse x2a2+y2b2=1 meet in the point (h,k). Then the mean position of the four points is

(a) (a2h2(a2+b2),b2k2(a2+b2))

(b) (a3h2(a2+b2),b3k2(a2+b2))

(c) (ah2(a2b2),bk2(a2b2))

(d) (a2h2(a2b2),b2k2(a2b2))

Show Answer Answer: (d)

10. The equation of the normal at the point (2,3) on the ellipse 9x2+16y2=180 is

(a) 3y=8x10

(b) 3y8x+7=0

(c) 8y+3x+7=0

(d) 3x+2y+7=0

Show Answer Answer: (b)

11. Number of distinct normal lines that can be drawn to the ellipse x2169+y225=1 from the point P (0,6) is

(a) one

(b) two

(c) three

(d) four

Show Answer Answer: (c)

12. Any ordinate MP of the ellipse x225+y29=1 meets the auxiliary circle at Q, then locus of the point of intersection of normals at P and Q to the respective curve is

(a) x2+y2=8

(b) x2+y2=34

(c) x2+y2=64

(d) x2+y2=15

Show Answer Answer: (c)

13. If the normals at P(θ) and Q(π2+θ) to the ellipse x2a2+y2 b2=1 meet the major axis at G and g respectively, then PG2+Qg2=

(a) b2(1e2)(2e2)

(b) a2(e4e2+2)

(c) a2(1+e2)(2+e2)

(d) b2(1+e2)(2+e2)

Show Answer Answer: (b)