prove that the locus of a centre of a circle which intercepts a chord of given length2aon the axis of - Maths - Conic Sections

Question

prove that the locus of a centre of a circle which intercepts a
chord of given length2aon the axis of x and passes through a given
point on the axis of y distant b from the origin is the curve
x2-2yb+b2 = a2 trace this parabola

Neha Sethi · Accepted Answer

Dear student
The equation of circle is:x-x12+y-y12=r2Here x-axis intercept=2a⇒2x12-x12+y12-r2=2aSquaring both sides, we getx12-x12+y12-r2=a2x12-x12-y12+r2=a2So, we get y12=r2-a2It passes through 0,bSo, we get 0-x12+b-y12=r2⇒x12+b-y12=r2⇒x12+b2+y12-2by1-r2=0⇒x12+b2+r2-a2-2by1-r2=0⇒x12+b2-2by1-a2=0So, locus of x1,y1 will be x2-2by+b2=a2Hence proved
For different values of a and b ,the  parabola is traced as:
<img alt="" src=
"https://s3mn%20mnimgs%20com/img/shared/content_ck_images/ck_5a0bea1d1c9fc%20png"
style="width: 1022px; height: 378px;">
Regards

prove that the locus of a centre of a circle which intercepts a chord of given length2aon the axis of x and passes through a given point on the axis of y distant b from the origin is the curve x2-2yb+b2 = a2 trace this parabola.

prove that the locus of a centre of a circle which intercepts a chord of given length2aon the axis of x and passes through a given point on the axis of y distant b from the origin is the curve x²-2yb+b² = a² trace this parabola.