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Let P be thepoint on the parabola, `y^2=8x`which is at a minimum distancefrom the centre C of thecircle,`x^2+(y+6)^2=1.`Then the equation of the circle, passing through C and having its centre at P is :(1) `x^2+y^2-4x+8y+12=0`(2) `x^2+y^2-x+4y-12=0`(3) `x^2+y^2-x/4+2y-24=0`(4) `x^2+y^2-4x+9y+18=0`A. `x^(2)+y^(2)-x+4y-12=0`B. `x^(2)+y^(2)=(1)/(4)x+2y-24=0`C. `x^(2)+y^(2)-4x+9y+12=0`D. `x^(2)+y^(2)-4x+8y+12=0` |
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Answer» Correct Answer - D Let `P(2t^(2), 4t)` be a point on `y^(2)=8x`. The coordinates of the centre C are(0, -6). `:." "CP^(2)=4t^(4)+(4t+6)^(2)` `"Let z =CP"^(2)` Then, `z=4t^(4)+4(2t+3)^(2)` `:." "(dz)/(dt)=16t^(3)+16(2t+3)" and "(d^(2)z)/(dt^(2))=48t^(2)+32` For maximum or minimum values of z, we must have `(dz)/(dt)=0rArr16(t^(3)+2t+3)=0` `rArr" "(t+1)(t^(2)-t+3)=0rArrt=-1` Clearly, `((d^(2)z)/(dt^(2)))_(t=-1)=80gt." So, z or CP is minimum when"` t=-1. So, the coordinates of P are (2, -4). The equation of the circle having centre at P(2, -4) adn passing through C(0, -6) is `(x-2)^(2)+(y+4)^(2)=(0-2)^(2)+(-6+4)^(2)` `"or, "x^(2)+y^(2)-4x+8y+12=0` ALITER Substituting t=2 in (i), the equation of the normal at P(4, 2) is y + 2x=12 whose x intercep is 6. The tangent to the circle at Q is perpendicular to the normal `y+2x==12`, So, the slope of the tangent is `1/2`. |
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