   Chapter 2.P, Problem 22P

Chapter
Section
Textbook Problem

# Given an ellipse x 2 / a 2 + y 2 / b 2 = 1 , where a ≠ b , find the equation of the set of all points from which there are two tangents to the curve whose slopes are (a) reciprocals and (b) negative reciprocals.

To determine

To find: The equation of the set of all points from which there are two tangents to the curve whose slopes are,

(a) Reciprocals

(b) Negative reciprocals

Explanation

1) Formula:

i. Point-slope formula: y-y1=m(x-x1)

ii.

a±b2=a2±2ab+b2

iii. Discriminant =  =b2-4ac

x=-b±b2-4ac2a

v. If m1 and m2 are roots of equation ax2+bx+c=0 then,

m1*m2=ca

2) Given:

Equation of ellipse is,

x2a2+y2b2=1

3) Calculation:

By using point-slope formula the equation of tangent which pass through the point (c, d) is,

y-d=m(x-c)

y=mx-c+d

To find the point where it touches the ellipse, substitute in equation of ellipse,

x2a2+(mx-c+d)2b2=1

By solving it,

x2b2+a2(mx-c+d)2a2b2=1

x2b2+a2(mx-c+d)2=a2b2

By using formula ii,

x2b2+a2[m2x-c2+2mx-cd+d2]=a2b2

Again by using formula ii,

x2b2+a2[m2(x2-2xc+c2)+2mx-cd+d2]=a2b2

By solving it,

x2b2+a2[m2x2-2xcm2+c2m2+2mxd-2mxcd+d2]=a2b2

x2b2+a2m2x2-2xcm2a2+c2m2a2+2mxda2-2mxcda2+d2a2=a2b2

x2b2+a2m2-2xcm2a2+2mxda2-2mxcda2+d2a2+c2m2a2-a2b2=0

x2b2+a2m2+x-2ca2m2+2ma2d+c2a2m2-2ma2cd+a2d2-a2b2=0

x2b2+a2m2+2mxa2d-cm+a2(c2m2-2mcd+d2-b2)=0

x2b2+a2m2+2ma2d-cmx+a2(d-cm2-b2)=0

Now, discriminant of this equation must be zero since a tangent line touches a curve at only one point

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