HYPERBOLA- 4 (Equation of Tangent)

Equation of Tangent

i. Point Form

x2a2y2b2=1

Differentiate w.r.t. x

2xa22ya2dydx=0dydx=b2xb2y

Slope of tangent =b2x1a2y1

Equation of tangent, yy1=b2x1a2y1(xx1)

xx1a2yy1 b2=x12a2y12 b2

xx1a2yy1 b2=1( But P(x1,y1) lies on hyperbola)

or T=xx1a2yy1 b21

Equation of tangent is T=0

Equation of tangent at point P(x1,y1) to the hyperbola (xh)2a2(yk)2 b2=1 is (xh)(x1h)a2(yk)(y1k)b2=1

ii. Parametric Form

Parametric equation of hyperbola is x=asecθ,y=btanθ

Equation of tangent is

xasecθybtanθ=1

iii. Slope Form

y=mx±a2m2b2

Hyperbola: (x2a2y2 b2=1) Point of contact.
Point form : xx1a2yy1 b2=1 (x1,y1)
Parametric Form : xasecθybtanθ=1 (asecθ,btanθ)
Slope form: y=mx±a2 m2b2 (±a2 ma2 m2b2,±b2a2 m2b2)

Examples

1. The equation of a tangent to the hyperbola 16x225y296x+100y356=0, which makes an angle π4 with the transverse axis, is

(a). y=x+2

(b). y=x5

(c). y=x+3

(d). x=y+2

Show Answer

Solution:

The equation of the hyperbola is

16(x26x)25(y24y)=356(x3)225(y2)216=1

The equation of tangent of slope m=tanπ4=1 to this hyperbola are

y2=1(x3)±25×116y2=x3±3y=x+2 or y=x4

Answer: (a)

2. The point of intersection of two tangents to the hyperbola x2a2y2b2=1, the product of whose slopes is c2, lies on the curve

(a). y2b2=c2(x2+a2)

(b). y2+a2=c2(x2b2)

(c). y2+b2=c2(x2a2)

(d). y2a2=c2(x2+b2)

Show Answer

Solution:

Let P(h,k) be the point of intersection of two tangents to the hyperbola x2a2y2 b2=1 The equation of tangent to hyperbola is

y=mx±a2 m2b2

If it passes through (h,k) then

k=mh±a2 m2b2

(kmh)2=a2 m2b2

m2( h2a2)2mkh+k2+b2=0

Let m1 and m2 be the slopes of the tangents passing throgh P. Then m1m2=k2+b2h2a2

c2=k2+b2 h2a2

Hence locus of P(h,k) is y2+b2=(x2a2)c2

Answer: (c)

3. If the tangents drawn from a point on the hyperbola x2y2=a2b2 to the ellipse x2a2+y2b2=1 make angles α and β with the transverse axis of the hyperbola, then

(a). tanαtanβ=1

(b). tanα+tanβ=1

(c). tanαtanβ=1

(d). tanαtanβ=1

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Solution:

Let P(h,k) be the point on the hyperbola x2y2=a2b2, then h2k2=a2b2.(i)

The equation of tangent to the ellipse is x2a2+y2b2=1 is

y=mx±a2m2+b2

If it passes through (h,k) then

k=mh±a2 m2+b2

(kmh)2=a2 m2+b2

m2( h2a2)2mkh+k2b2=0

Let m1 and m2 be the roots of this equation, then

m1m2=k2b2 h2a2=1 (from equation (i))

tanαtanβ=1

Answer: (c)

4. If the line 2x+6y=2 touches the hyperbola x22y2=4, then the point of contact is

(a). (2,6)

(b). (5,26)

(c). (12,16)

(d). (4,6)

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Solution:

Equation of hyperbola is x2(2)2y2(2)2=1

Equation of tangent is y=23x+23

m=23,a=2,b=2,c=23 and c2=a2m2b2

Point of contact is (±a2ma2m2b2,±b2a2m2b2) i.e. (4,±6)

Answer: (d)

5. Number of real tangents can be drawn from the point (5,0) to the hyperbola x216y29=1 is

(a). 2

(b). 1

(c). 0

(d). 4

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Solution:

S1=2516091>0

Point (5,0) lies inside the hyperbola. Hence no tangents can be drawn.

Answer: (c)

Practice questions

1. P is a point on the hyperbola x2a2y2 b2=1, N is the foot of the perpendicuilar from P on the transverse axis. The tangent to the hyperbola at P meets the transverse axis at T. If O is the centre of the hyperbola, then OT×ON is

(a). b2

(b). a2

(c). ab

(d). none of these

Show Answer Answer: (b)

2. Tangents drawn from the point (c,d) to the hyperbola x2a2y2b2=1 make angles α and β with the x-axis. If tanαtanβ=1, then c2d2=

(a). a2b2

(b). a2b2

(c). a2+b2

(d). none of these

Show Answer Answer: (c)

3. The common tangents to the two hyperbolas x2a2y2 b2=1 and y2a2x2 b2=1 is

(a). y=x+a2b2

(b). y=x+a2+b2

(c). y=2x+a2b2

(d). none of these

Show Answer Answer: (a)

4. The equation of the tangent to the curve 4x29y2=1 which is parallel to 4y=5x+7 is

(a). 30x+24y=720

(b). 7y=6x15

(c). y=327x+157

(d). none of these

Show Answer Answer: (d)

5. The locus of a point P(h,k) moving under the condition that the line y=hx+k is a tangent to the hyperbola x2a2y2b2=1 is

(a). Parabola

(b). circle

(c). ellipse

(d). hyperbola

Show Answer Answer: (d)