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If the locus of the point from where two of three normals to the curve y^2+2 y-4 x+5=0 are perpendicular is (y- k 2 )^2-(x- )=0 , then find the value of -2 k .

Correct answer

8

Step-by-step solution

The curve is (y+1)^2=4(x-1) equation of the normal to the given curve is aligned & y +1= m ( x -1)-2 ~m - m ^3 & y = mx -3 ~m - m ^3-1 aligned which passes through (h, k) aligned & m₁ m₂ m₃=-(1+k) & m₃=(1+k) m₁ m₂=-1 & (1+k)^3+(1+k)(3-h)+(1+k)=0 & (1+k)^2+3-h+1=0 & (y+1)^2=x-4 aligned locus of ( h , k ) are aligned & C ₁:( y +1)^2= x -4 & k 2 =-1 k =-2, =4 & -2 k =8 aligned

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