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