#### 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

#### Answer

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

(a) Reciprocals is, x2-y2=a2-b2

(b) Negative reciprocals is, x2+y2=a2+b2

#### Explanation

**1) Formula:**

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

ii.

a±b2=a2±2ab+b2

iii. Discriminant = ∆ =b2-4ac

iv. Quadratic formula:

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.

That is, b2-4ac=0

⇒2ma2d-cm2-4b2+a2m2a2d-cm2-b2=0

⇒4m2a4d-cm2=4b2+a2m2a2d-cm2-b2

Divide by 4a2

⇒m2a2d-cm2=b2+a2m2d-cm2-b2

⇒m2a2d-cm2=b2+a2m2d-cm2-b2b2+a2m2

⇒b2b2+a2m2=b2+a2m2d-cm2-m2a2d-cm2

⇒b2b2+a2m2=d-cm2(b2+a2m2-m2a2)

⇒b2b2+a2m2=d-cm2(b2)

Divide by b2

⇒b2+a2m2=d-cm2

By using formula ii,

⇒b2+a2m2=d2-2cdm+c2m2

By rearranging the terms,

⇒m2a2-c2+2cdm+b2-d2=0

Now by using quadratic formula,

m=-2cd±2cd2-4a2-c2b2-d22(a2-c2)

It has two solutions m1 and m2

By using formula, and we can see that

m1*m2=b2-d2a2-c2

a) Slopes are reciprocals

That is m1=m and m2=1m

Therefore,

m1*m2=1=b2-d2a2-c2

⇒b2-d2=a2-c2

Replace (c, d) with (x, y)

⇒b2-y2=a2-x2

⇒x2-y2=a2-b2

b) Slopes are negative reciprocals

That is m1=m and m2=-1m

Therefore,

m1*m2=-1=b2-d2a2-c2

⇒b2-d2=c2-a2

Replace (c, d) with (x, y)

⇒b2-y2=x2-a2

⇒x2+y2=a2+b2

**Conclusion:**

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

(a) Reciprocals is, x2-y2=a2-b2

(b) Negative reciprocals is, x2+y2=a2+b2