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Question 89 of 126

Q.The locus of the foot of perpendicular from the focus on any tangent to the hyperbola x2a2−y2b2=1\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1 is :

(a) x2+y2=a2+b2x^2 + y^2 = a^2 + b^2
(b) x2+y2=a2−b2x^2 + y^2 = a^2 - b^2
(c) x=0x = 0
(d) x2+y2=a2x^2 + y^2 = a^2
Puducherry TnboardTamil Nadu HSC (DGE) Board 2018MCQ· 1mImportance★★★★★
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Computing the foot of the perpendicular from a focus of the hyperbola onto a general tangent line shows the locus is the auxiliary circle x2+y2=a2x^2+y^2=a^2.

  1. A tangent to x2a2−y2b2=1\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1 in slope form is y=mx+cy=mx+c with the tangency condition c2=a2m2−b2c^2=a^2m^2-b^2, i.e. mx−y+c=0mx-y+c=0.
  2. Take the focus S=(ae,0)S=(ae,0), where b2=a2(e2−1)b^2=a^2(e^2-1).
  3. Using the perpendicular-foot formula for a point (x0,y0)(x_0,y_0) onto ux+vy+w=0ux+vy+w=0 (here u=m, v=−1, w=cu=m,\,v=-1,\,w=c): with k=ame+cm2+1k=\dfrac{ame+c}{m^2+1}, the foot is x=ae−mk, y=kx=ae-mk,\ y=k.
  4. Compute x2+y2=(ae−mk)2+k2=a2e2−2aemk+k2(m2+1)x^2+y^2=(ae-mk)^2+k^2 = a^2e^2-2aemk+k^2(m^2+1).
  5. Since k(m2+1)=ame+ck(m^2+1)=ame+c, this simplifies to a2e2+k(c−ame)a^2e^2 + k(c-ame), and substituting k=ame+cm2+1k=\dfrac{ame+c}{m^2+1} gives x2+y2=a2e2+c2−a2m2e2m2+1x^2+y^2=a^2e^2+\dfrac{c^2-a^2m^2e^2}{m^2+1}. …

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