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The normal to the curve x^2=4ypassing (...

The normal to the curve `x^2=4y`passing (1,2) is(A) `x + y = 3` (B) `x y = 3` (C) `x + y = 1` (D) `x y = 1`

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Given, `x^2=4y `
Slope of tangent to the curve will be` 4(dy/dx)​=2x`
⇒`dy/dx​=2x`
​ So slope of normal will be =`−dx/dy​=-2/x`
​ Let the point on the curve be (h,k)
So slope of normal will be `-2/h​`
Equation of normal is given by `(y−k)=-2/h​(x−h)`
...
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